## UVA Problem 10432 – Polygon Inside A Circle Solution

UVA Problem 10432 – Polygon Inside A Circle Solution:

Solving Technique:

This is a rather easy geometry / computational geometry problem. Given the radius of a Circumscribed circle and count of sides of a polygon the task is to find the area of the polygon. A Circumscribed circle is a circle that passes through all vertices of a plane figure and contains the entire figure in its interior.

The formula below can be written into a single formula by combining all the formulas. More information and the combined formula can be found here.

##### Visual Explanation:

I have tried to explain the concept below using figures. They are not drawn to scale. The small circles represent intersection point between polygon vertices and the circumscribed circle.

##### Example:

It is an example with radius, r = 2 and sides, n = 8.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

2 2000
10 3000


Output:

12.566
314.159


### Code:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 10432 - Polygon Inside A Circle
* Technique: circumcircle Or, Isocele Area calculation.
*/

#include<stdio.h>
#include<string.h>
#include<math.h>

int main(){

//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

double r;
int n;

while( scanf( "%lf%d", &r, &n ) == 2 ){

// Angle between each two points for every point.
double PHI = ( double ) 360 / n ;

// For each Isosceles in the polygon the angle between the base and radius.
double THETA = (double) 90 - ( PHI / 2 );

// Convert Degree angle to Radian to use in code.
double THETA_RADIAN = THETA * M_PI / 180;

//  a is base.
double a = 2 * r * cos( THETA_RADIAN );

// H is the height.
double h = r * sin( THETA_RADIAN );

// S represent Area of a single segment.
double S = (a * h) / 2;

// S * n is the are of complete polygon.
printf("%.3lf\n",  S * n );

}

return 0;
}


## UVA Problem 10106 – Product Solution (Lattice Multiplication)

UVA Problem 10106 – Product Solution:

Solving Technique:

This problem is quite simple. Given two very large integers multiply them and print them. By large I mean almost 251 characters long for this problem. So string or character array must be used.

This problem can solved in many ways Big Integer, FFT, Grade School etc. But I have implemented lattice multiplication / Chinese multiplication to solve this problem.

Code Explanation:

The code below requires quite a bit of polish to understand easily ( A lot other post require the same. Maybe I will update this post sometimes in future or not. ).

This code is not quite space efficient. I have used several arrays to make it easy to understand but all of the tasks below can be done using the large array. Also the running time can further be improved to $O(m*n)$ instead of $O(n^2)$. Where, m and n are length of two arrays and they may not be equal in size.

I have provided some graphical representations below to make it easier to understand.

This is an example of how the multiplication works in this code. As mentioned above the process can be sped up using rectangular matrix instead of square if the length is not equal.

This is a representation of how the table / square matrix is filled for further processing.

Multiplication starts with the last character of string 1 and proceeds to first character. For string 2 multiplication starts from the first character till last character. This way each character from string 1 and string 2 is first converted to a digit, then they are multiplied. If the result is two digits it is divided by 10.

The remainder is written as the lower left portion ( indicated by lower in the structure ) and the quotient is written as the upper left portion ( indicated by upper in the structure ).

This is graphical representation of how the structure is traversed. If you have trouble understanding how the recursive function traversing the structure, then take a look at UVA problem 264 – Count the Cantor Solution. The traversal is somewhat similar to that and I have provided explanation int that post.

Rest is explained in the code comments.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

12
12
2
222222222222222222222222


Output:

144
444444444444444444444444


### Code:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   Lattice Multiplication.
* Technique: High precision large number multiplication.
*            Structure array representing upper left and
*            and lower right corner in single cell.
*/

#include<stdio.h>
#include<string.h>

#define M 260

int N;

// Assuming both strings are of same length.
struct multiplicationTable{
int upper;
int lower;
};
struct multiplicationTable node[M][M];

static char string1[M];
static char string2[M];

// stores the diagonal sums;
int sum;

// decides whether to get the upper or lower
// value based on even or odd.
int m;

int recursiveAdder( int i, int j ){

// Termination condition.
if( j < 0 || i >= N )
return sum;

int val;

// Whether to get the upper left corner or
// the lower right corner.
if( m % 2 ){
val = node[i][j].upper;
j = j - 1;
}
else{
val = node[i][j].lower;
i = i + 1;
}
++m;

// store the sum of the whole diagonal.
sum = sum + val;

// recursively visit all row ans column
// diagonally ( at least on pen and paper format ).
// actually moves more like the snake in the snakes game.

return sum;
}

// Debug.
// Print the matrix showing the multiplications.
// Please note left and right directions may be different.
void printMultiplicationMatrix(){

printf("\n\n");
for( int i = 0; i < N; ++i ){
for( int j = N - 1; j >= 0; --j )
printf("%d,%d, ", node[i][j].upper, node[i][j].lower );
printf("\n");
}

}

int main(){

//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

while( gets(string1) ){
gets(string2);

int len1 = strlen(string1);
int len2 = strlen(string2);

// Fix length if both string are not equal size.
if( len1 > len2 ){
N = len1;

int shiftWidth = len1 - len2;

for( int i = len1 - 1; i >= 0; --i )
string2[i + shiftWidth] = string2[i];

for(int j = 0; j < shiftWidth; ++j)
string2[j] = '0';

}
else if(len2 > len1){
N = len2;

int shiftWidth = len2 - len1;

for( int i = len2 - 1; i >= 0; --i )
string1[i + shiftWidth] = string1[i];

for(int j = 0; j < shiftWidth; ++j)
string1[j] = '0';

}
else N = len1;

//printf("%s \n%s \n", string1, string2);

int k = N - 1;

// Multiply the numbers digit by digit and set in the lattice.
for( int i = 0; string2[i]; ++i ){
for( int j = 0; string2[j]; ++j ){

int num1 = string1[k] - '0';
int num2 = string2[j] - '0';

int multiply = num1 * num2;

node[j][k].upper = multiply / 10;
node[j][k].lower = multiply % 10;

}
--k;
}

//printMultiplicationMatrix();

// Lattice is divided into two parts upper left half and
// lower right half.

// result of upper half
int upperHalfResult[N];

int i = N - 1;
for(; i >= 0; --i){
sum = 0;
m = 1;
}

// result of upper half
int lowerHalfResult[N];

i = 0;
for(; i < N; ++i){
sum = 0;
m = 0;
lowerHalfResult[i] = recursiveAdder(i, N - 1);
}

// Combine upper and lower left half to a single array to fix addition
// problems.
int combinedRawResult[N + N];
i = 0;
for(; i < N; ++i )
combinedRawResult[i] = upperHalfResult[i];
for(k = 0; i < N + N; ++i, ++k )
combinedRawResult[i] = lowerHalfResult[k];

// If a cell has more than 9 then it should be added to the next cell.
for( int i = N + N - 1; i >= 0; --i ){

if( combinedRawResult[i] > 9 ){
combinedRawResult[i - 1] = combinedRawResult[i - 1] + combinedRawResult[i] / 10;
combinedRawResult[i] = combinedRawResult[i] % 10;
}

}

for(i = 0; i < N + N; ++i)
if(combinedRawResult[i]) break;

// print if the result can be printed or its zero.
bool zero = true;
for(; i < N + N; ++i){
printf("%d", combinedRawResult[i]);
zero = false;
}

// If the result is zero.
if( zero )
printf("0");

printf("\n");

}

return 0;
}


## UVA Problem 11988 – Broken Keyboard (a.k.a. Beiju Text) Solution

UVA Problem 11988 – Broken Keyboard (a.k.a. Beiju Text) Solution:

Solving Technique:

The input may be classified into three things. One is append to beginning represented with ‘[‘, another is append to end represented with ‘]’ and the other types any other character except above mentioned. The other character are appended to the next position of last inserted character.

So task is move all characters following ‘[‘ to beginning until ‘]’ is found or end of string is reached. Similar logic applies to ‘]’ character. Also it is a recursive definition, keep applying this logic for all the following third brackets.

###### Other ideas:

Some other ideas to solve this problem can be, formatting the input to remove useless brackets then create a tree from inputs and perform in-order traversal to get the result. Another idea is create a linked list of strings.

###### Code Explanation:

I may later ( takes quite bit of time to create graphics ) provide a graphical representation of inserting in the doubly linked list ( 2nd code ). Meanwhile I provided a commented version ( 2nd code ) to make it a little easier to understand.

Here I have provided two codes. Both are linked list implementations. First one is singly linked list with no tail pointer( rather inefficient although I try to reduce some complexity by Boolean flag checking to reduce traversing the list to find tail pointer ). It is not able to keep track of tail pointer efficiently. Need to do some traversing ( some times almost the whole linked list ) to find the tail pointer.

The second implementation is doubly linked list with tail node as the tail pointer. The tail pointer has character has 0 which is ASCII decimal code for NULL character. I use this to stop printing otherwise a space character is printed at the end of output.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University


Output:

BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University


### Code Singly Linked list Without Tail:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 11988 - broken keyboard
* Technique: Singly Linked List with no
*            tail Pointer(slow).
*/

#include<stdio.h>
#include<string.h>

#define N 100000

static char input[N];

struct vertex{
char c;
struct vertex* next;
};

typedef struct vertex node;

node* insertNode(node* n, char c){

node* temp = new node();
temp->c = c;
temp->next = n->next;
n->next = temp;

return temp;
}

void printList( node* dummy ){

node* tmp = dummy;

while( tmp = tmp->next )
putchar( tmp->c );

printf("\n");
}

int main(){

//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

while( gets(input) ){

node* dummy = new node();
dummy->next = NULL;
node* cur = dummy;

node* tail = dummy;

bool rightopen = false;

for(int i = 0; input[i]; ++i){

if( input[i] == '[' ){
cur = dummy;
rightopen = true;
continue;
}
if( input[i] == ']' ){
rightopen = false;

node *tmp = tail;

while( tmp->next != NULL )
tmp = tmp->next;

tail = cur = tmp;
continue;
}

cur = insertNode(cur, input[i] );
if( !rightopen )
tail = cur;
}

printList( dummy );

}

return 0;
}


### Code Doubly Linked list With Tail Pointer:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 11988 - broken keyboard
* Technique: Doubly Linked List with
*            dummy tail node as Pointer.
*/

#include<stdio.h>
#include<string.h>

#define N 100000

static char input[N];

struct vertex{
char c;
struct vertex* next;
struct vertex* prev;
};
typedef struct vertex node;

node* insertNode(node* n, char c){

// Create the new node to hold current character.
node* temp = new node();
temp->c = c;

// Create forward connection from this created node
// to the next node of passed in node (n).
// Also create backward connection from the passed
// in node to current node.
temp->next = n->next;
n->next->prev = temp;

// Create forward connection from passed in node (n) to the
// newly created node (temp).
// Also create backward connection from created node to the
// passed in node.
n->next = temp;
temp->prev = n;

// return the pointer to the current node.
return temp;
}

void printList( node* dummy ){

// Dummy (head) is not a part of output.
node* tmp = dummy->next;

// Since I use tail pointer, check if its tail.
while( tmp->c ){
putchar( tmp->c );
tmp = tmp->next;
}

printf("\n");
}

int main(){

//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

while( gets(input) ){

// Dummy node is just to keep track of beginning.
// It does not contain any input character.
// Tail is used to insert before the last node.
// Tail is also another dummy node keeps track of
// the end.
node* dummy = new node();
node* tail = new node();

// NULL may not be necessary since I use tail character
// checking to stop printing.
tail->prev = dummy;
dummy->next = tail;
tail->next = NULL;
dummy->prev = NULL;

// Set current node to dummy (beginning). Set tail
// to NULL character equivalent ASCII value.
node* cur = dummy;
tail->c = 0;

for(int i = 0; input[i]; ++i){

// Update node pointer to beginning based on bracket.
// If not bracket use previous pointer location.
if( input[i] == '[' ){
cur = dummy;
continue;
}
if( input[i] == ']' ){
cur = tail->prev;
continue;
}

cur = insertNode(cur, input[i] );
}

// Print output characters one by one.
printList( dummy );

}

return 0;
}


## UVA Problem 10189 – Minesweeper Solution

UVA Problem 10189 – Minesweeper Solution:

Solving Technique:

Given a mine field that is a matrix / 2D array, produce an output that contains count of adjacent mines for each squares.

In a 2D array for each squares there are at most adjacent 8 squares. If the current position is i and j then,

$\begin{bmatrix} (i-1,j-1) & (i-1,j) & (i-1,j+1) \\ (i+0,j-1) & (i,j) & (i+0,j+1) \\ (i+1,j-1) & (i-1,j) & (i+1,j+1) \\ \end{bmatrix}$

Just traverse the matrix row-column wise and check its adjacent squares for getting mine count for current position. The adjacent squares check can be implemented with 8 if conditions for each one.

###### Here the arrow from the center shows where checking starts. The 1D arrays drow and dcol hold the row and column values the way shown above. It can be changed by modifying drow and dcol arrays.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0


Output:

Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100


### Code Using Multiple if Conditions:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 10189 - Minesweeper
* Technique: 2D Array / Matrix Boundary checking using
*            if conditions.
*/

#include<stdio.h>
#include<string.h>

#define MAXSIZE 101

static char MineField[MAXSIZE][MAXSIZE];

int main(){

//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

int n, m;

int FieldNumber = 0;

while( scanf("%d%d", &n, &m), n ){

getchar();

for(int i = 0; i < n; ++i)
scanf("%s", &MineField[i]);

if( FieldNumber )
printf("\n");

for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j){

if( MineField[i][j] == '*' )
continue;

int temp = 0;

if( i + 1 < n && MineField[i + 1][j] == '*' )
++temp;
if( i + 1 < n && j + 1 < m && MineField[i + 1][j + 1] == '*' )
++temp;
if( j + 1 < m && MineField[i][j + 1] == '*' )
++temp;
if( i - 1 >= 0 && j + 1 < m && MineField[i - 1][j + 1] == '*' )
++temp;
if( i - 1 >= 0 && MineField[i - 1][j] == '*' )
++temp;
if( i - 1 >= 0 && j - 1 >= 0 && MineField[i - 1][j - 1] == '*' )
++temp;
if( j - 1 >= 0 && MineField[i][j - 1] == '*' )
++temp;
if( i + 1 < n && j - 1 >= 0 && MineField[i + 1][j - 1] == '*' )
++temp;

MineField[i][j] = temp + '0';

}
}

printf("Field #%d:\n", ++FieldNumber);

for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j)
putchar(MineField[i][j]);
printf("\n");
}

}

return 0;
}


### Code Bound checking using Array & for Loop:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 10189 - Minesweeper
* Technique: 2D Array / Matrix Boundary checking using
*            co-ordinate array and for loop.
*/

#include<stdio.h>
#include<string.h>

#define MAXSIZE 101

static char MineField[MAXSIZE][MAXSIZE];

// Co-ordinates / directions of adjacent 8 squares.
// W, SW, S, SE, E, NE, N, NW
static const int drow[] = {0, 1, 1, 1, 0, -1, -1, -1};
static const int dcol[] = {-1, -1, 0, 1, 1, 1, 0, -1};

int main(){

//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

int n, m;

int FieldNumber = 0;

while( scanf("%d%d", &n, &m), n ){

getchar();

for(int i = 0; i < n; ++i)
scanf("%s", &MineField[i]);

if( FieldNumber )
printf("\n");

for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j){

int temp = 0;

// If mine found do nothing.
if( MineField[i][j] == '*' )
continue;

// For each adjacent squares of the current square calculate mine count.
// and set the count in current square.
for(int k = 0; k < 8; ++k){

// Check if out of bound of the 2D array or matrix.
if( i + drow[k] < 0 || j + dcol[k] < 0 || i + drow[k] >= n || j + dcol[k] >= m )
continue;

// Check the appropriate co-ordinate for mine, if mine found increase count.
if( MineField[i + drow[k] ][j + dcol[k]] == '*' )
++temp;

}

// All adjacent squares checked set the mine count for current squares.
MineField[i][j] = temp + '0';

}
}

printf("Field #%d:\n", ++FieldNumber);

for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j)
putchar(MineField[i][j]);
printf("\n");
}

}

return 0;
}


## UVA Problem 11565 – Simple Equations Solution

UVA Problem 11565 – Simple Equations Solution:

## UPDATE:

As a commenter below pointed out the solution is wrong due to the fact I made some wrong assumptions. Kindly check other sources for correct solution. This was accepted due to weak test cases. Also check udebug for different test cases.

Problem Statement Description:

Given input of $A, B, C$ where $0 \leqslant A, B, C \leqslant 10000$ and three equations,

$\mathbf{ x + y + z = A }$
$\mathbf{ x.y.z = B }$
$\mathbf{ x^2 + y^2 + z^2 = C }$

The task is to find values of x, y, z where x, y, z are three different integers which is ( x != y != z ). Meaning x can not be equal to y, t can not be equal to z and z can not be equal to x.

The problem asks to output the least value of x then y then z meaning the output will be in x < y < z order.

#### Things to notice:

One very important thing to notice is that for all three equation the values of x, y, z can be interchanged and it will still satisfy the problem.

For example,
$\mathbf{ if, x = 5, y = 0, z = 9 }$

then it can also be written as,
$\mathbf{ x = 0, y = 5, z = 9 }$
$\mathbf{ x = 9, y = 5, z = 0 }$

etc. all possible combinations. But the problem asks for least value of x then y then z.

So if after calculation the result is x = 9, y = 0, z = 5 which is supposed to be x = 0, y = 5, z = 9. In that case the values can be sorted which will result in x = 0, y = 5, z = 9 and this still satisfies the equations.

#### Solution Technique Explanation:

The simple approach to solve this problem is try all possible combinations of x, y, z and see if it produces a correct result.

The values of x, y, z being negative also satisfies this problem. Forgetting the value of negatives for the moment and focusing on the positive values. To get the naive loop count limit see x or y or z can be at least 0 and at most 100.

Same goes for the negative portion so looping for -100 to 100 for each 3 nested loops seem easy ( I haven’t given the 3 nested loop solution here. It can be found on other sites ). So from -100 to 0 to 100 there are 201 number in between. So naive solution requires 201*201*201 which is more than 8 million (8120601). This program should do 117 * 45 = 5265 iterations per test case.

###### This problem can be solve much faster using some algebraic substitution. Notice,

$\mathbf{ x + y + z = A ........ (i) \\ }$
$\mathbf{ x.y.z = B ........ (ii) \\ }$
$\mathbf{ x^2 + y^2 + z^2 = C ........ (iii) \\ }$

from equation (ii),

$\mathbf{ x.y.z = B }$

This can be rewritten as,

$\mathbf{ z = \frac{B}{x.y} ........ (iv) }$

from equation (i),

$\mathbf{ x + y + z = A }$

substituting value of z from (iv) to (i) gives,

$\mathbf{ x + y + \frac{B}{x.y} = A }$

This can be rewritten as,

$\mathbf{ \frac{B}{x.y} = A - x - y ........ (v) }$

similarly from (iii) and (iv),

$\mathbf{ x^2 + y^2 + z^2 = C }$

plugging in the value of z and solving for x and y,

$\mathbf{ x^2 + y^2 + ( \frac{B}{x.y} )^2 = C ........ (vi) }$

Now from (v) and (vi),
$\mathbf{ x^2 + y^2 + (A - x - y)^2 = C }$

Notice this equation above does not contain z. Now z can be calculated in terms of B, x, y.

This will completely remove one of the nested loops. Further improvements can be made by cutting down number of loop iterations.

###### Important point to notice that A, B, C can be at most 10000. Also x, y, z can differ by 1 and satisfy the equation. From above equations,

$\mathbf{ x + y + z = A ........ (i) }$
$\mathbf{ x.y.z = B ........ (ii) }$
$\mathbf{ x^2 + y^2 + z^2 = C ........ (iii) }$

So if x, y, z differ by 1 then,

$\mathbf{ y = x + 1 }$
$\mathbf{ z = x + 2 }$

from equation (i),

$\mathbf{ x + (x + 1) + (x + 2) = A }$

but A can be at most 10000.

$\mathbf{ x + (x + 1) + (x + 2) = 10000 }$
$\mathbf{ 3x + 3 = 10000 }$

So, x = 3332.3333….

But to keep it simple, notice the lower order term 3 does not make much difference. By setting x = y = z the equation can be,

$\mathbf{ x + x + x = 10000 }$
$\mathbf{ 3x = 10000 }$
$\mathbf{ x = 3333.3333.... }$

###### from equation (iii) since x is squared so it to get its iteration just calculate the square root. or, setting x = y = z and B = 10000 in equation (iii) gives,

$\mathbf{ x^2 + x^2 + x^2 = 10000 }$
$\mathbf{ 3x^2 = 10000 }$
$\mathbf{ x = 57.735.... }$

after rounding of,
x = 58

So the loop range for x is [-58…58].

###### Again from equation (ii) the iteration limit for y can be counted using x = y = z,

$\mathbf{ y.y.y = 10000 }$
$\mathbf{ y^3 = 10000 }$
$\mathbf{ y = 21.544.... }$

Rounded of to,
y = 22

So iteration limit for y will be from [-22…22].

Calculating using y + 1 result in rounded range [-21…21]. Although I have not tested it.

###### Doing further algebraic simplification will remove another loop as z can be calculated in terms of A,B,C using,

$\mathbf{ (A-z)^2 - 2.\frac{B}{z} = C - z^2 }$

where,

$\mathbf{ xy = \frac{B}{z} }$

and,

$\mathbf{ x + y = A - z }$

using the main equation above, the value of z can be found and by solving the other equations the values of x and y can be calculated.

Also the loops can be input dependent. If max of A, B, C is bigger than 58 than choose 58 other wise choose the max as the loop counter.

There may be more improvements possible but that’s all I can come up with now. One final improvement can be made by breaking from all loops at once once the values are found. Although goto isn’t recommended c++ doesn’t support label to break out of multiple loops unlike java.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

2
1 2 3
6 6 14

Output:

No solution.
1 2 3

### Optimized Code without goto:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 11565 - Simple Equations
* Technique: Mathematical elimination and substitution
*/

#include<cstdio>
#include<algorithm>
#include<iostream>

int main(){
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);

int n;
scanf("%d", &n);

int A, B, C;

while( n-- ){
scanf("%d%d%d", &A, &B, &C);

bool solutionFound = false;
int x, y, z;

for( x = -58; x <= 58; ++x ){
for( y = -22; y <= 22; ++y ){

if( x != y && ( (x * x + y * y) + (A - x - y) * (A - x - y) == C )  ){

int temp = x * y;

if( temp == 0 ) continue;

z = B / temp;

if( z != x && z != y && x + y + z == A   ){
if(!solutionFound){
int tmpArray[3] = {x, y, z};
std::sort(tmpArray, tmpArray + 3);
std::cout << tmpArray[0] << " " << tmpArray[1] << " " << tmpArray[2] << "\n";

solutionFound = true;
break;
}
}

}
}
}

if(!solutionFound) std::cout << "No solution." << "\n";

}

return 0;
}


### Optimized Code without goto:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 11565 - Simple Equations
* Technique: Mathematical elimination and substitution
*/

#include<cstdio>
#include<algorithm>
#include<iostream>

int main(){
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);

int n;
scanf("%d", &n);

int A, B, C;

while( n-- ){
scanf("%d%d%d", &A, &B, &C);

bool solutionFound = false;
int x, y, z;

for( x = -58; x <= 58; ++x ){
for( y = -22; y <= 22; ++y ){

if( x != y && ( (x * x + y * y) + (A - x - y) * (A - x - y) == C )  ){

int temp = x * y;

if( temp == 0 ) continue;

z = B / temp;

if( z != x && z != y && x + y + z == A   ){
if(!solutionFound){
int tmpArray[3] = {x, y, z};
std::sort(tmpArray, tmpArray + 3);
std::cout << tmpArray[0] << " " << tmpArray[1] << " " << tmpArray[2] << "\n";

solutionFound = true;
goto done;
}
}

}
}
}

if(!solutionFound) std::cout << "No solution." << "\n";
done:;

}

return 0;
}


## UVA Problem 264 – Count on Cantor Solution

UVA Problem 264 – Count on Cantor Solution:

Solving Technique:

Warning this is a terrible dynamic programming and memoization solution. If you want a fast and efficient solution then, this isn’t what you are looking for.

#### Problem Statement:

Given an input integer that represent a rational number sequence term shown by cantor, output the rational number for that term.

##### Solution Types:

I have given two solution, one of which uses a 2D array to store the rational number sequence as shown by cantor and using a traversal pattern memoize the series in another array. Although the first one can be improved to only traverse 1/4 th of the array but that’s still a lot.

The next solution discards the 2D array and using the traversal pattern with some optimization memoizes the series.

###### The time difference as of UVA submission running time is,

0.245 ms (1st technique)
0.072 ms (2nd technique)

Removing the recursion by replacing with iterative version ( A bit harder to do ) will make it even faster. I think this one can be further improved but have no idea how.

#### Figuring out the algorithm:

Note everything here is assuming indexing starts from 1 instead of 0. But in my code I start with index 0 instead of 1.

##### Points To Notice:

1. Moves right when row number is 0 and column number is even.
2. Moves down-left when row number is 0 and column number is odd.
3. Moves down when column number is 0 and column number is odd.
4. Moves up-right when column number is 0 and column number is even.
5. Diagonal traversal gets bigger by 1 unit for each down or right movements.

##### Traversing the Matrix:

$Traverse(r,c,index,diagonal) = \begin{cases} \rightarrow \text{(1 times)} & \forall : row = 0; col \in Even \\ \swarrow \text{(c - 1 times)} & \forall : row = 0; col \in Odd \\ \downarrow \ \ \text{(1 times)} & \forall : col = 0; row \in Odd \\ \nearrow \text{(r - 1 times)} & \forall : col = 0; row \in Even \\ \end{cases}$

OR, In my code,

$Traverse(r,c,index,diagonal) = \begin{cases} \rightarrow \text{(1 times)} & \forall : row = 0; col \in Even \\ \swarrow \text{(c + diagonal times)} & \forall : row = 0; col \in Odd \\ \downarrow \ \ \text{(1 times)} & \forall : col = 0; row \in Odd \\ \nearrow \text{(r + diagonal times)} & \forall : col = 0; row \in Even \\ \end{cases}$

Note: Increment the index and diagonal also.

##### Filling the Matrix (1st code only):

$CantorTable_{i,j} = \begin{cases} numerator = row, & \forall : Column, rows \\ denominator = column, & \forall : Column, rows \\ \end{cases}$

##### Base case:

Note when row or the column is equal to 0 then there is a turn. If row or column becomes less than 0 then it should stop.

Also the amount values to memoize is N which is the maximum cantor sequence value for this problem. So if N values are memoized then the process should stop.

if (r < 0 || c < 0 || index > M)
return;


##### Storing the fractions in 2D Struct Array:

Instead of storing the values in string it is better to store the fraction in a struct. There are 3 things in the fraction the numerator, a division sign , the denominator. There is no need to store the division sign since all elements within matrix are fraction.

struct CantorSequence{
int numerator;
int denominator;
};


The first solution requires a total of $O( M + \frac{N(N+1)}{2} )$ space including storing the cantor sequence in another array and pre calculating the 2D cantor table and traversing.

The second solution requires a total of $O( M )$ space including storing the cantor sequence in another array.

Here, M is 10000000 and N is 4500. N can be adjusted but 4500 is a safe value.

###### Please point out if the post contains mistakes.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

3
14
7
10000000


Output:

TERM 3 IS 2/1
TERM 14 IS 2/4
TERM 7 IS 1/4
TERM 10000000 IS 2844/1629


### Code Recursive DP 2D struct Traversal and Memoize:

/***********************************************************************
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 264 - Count on Cantor
*
* Technique: Zig Zag / Spiral Diagonal traversal,
*            2D struct array,
*            2D half diagonal fill,
*            Recursive Dynamic Programming
***********************************************************************/

#include<stdio.h>
#include<string.h>

// N * N should be almost twice greater than or equal to M.
// Or, N should be such that N*(N+1)/2 is greater than M.
#define N 4500

#define M 10000000

/**
* Table to construct the sequence
*/
struct CantorTable{
int numerator;
int denominator;
} CT[N][N];

// Holds cantor values from 1 to M by index
CantorTable OrderedCantor[M];

/**
* Fill in the cantor table.
*/
void CantorFill(){

for(int row = 0; row < N; ++row){

int cutoff = N - row;

for(int col = 0; col < cutoff; ++col){
CT[row][col].numerator   = row + 1;
CT[row][col].denominator = col + 1;
}

}

}

void RecursiveCantor(int r, int c, int index, int diagonal){

// base case
if( r < 0 || c < 0 || index > M ){
return;
}

OrderedCantor[index].numerator = CT[r][c].numerator;
OrderedCantor[index].denominator = CT[r][c].denominator;

// Amount of times to travel in diagonal
++diagonal;

if(r == 0){

// when odd move down left
if(c % 2){
for(int i = 0; i < c + diagonal; ++i){
++index;
r = r + 1;
c = c - 1;
RecursiveCantor( r, c, index, diagonal );
}
}

// when even Move right
else{
++index;
c = c + 1;
RecursiveCantor( r, c, index, diagonal );
}
}

if(c == 0){

// when odd move down
if(r % 2){
++index;
r = r + 1;
RecursiveCantor( r, c, index, diagonal );
}

// when even Move up right
else{
for(int i = 0; i < r + diagonal; ++i){
++index;
r = r - 1;
c = c + 1;
RecursiveCantor( r, c, index, diagonal );
}
}

}

}

int main() {

// Initialize Cantor table
CantorFill();

// Pass in row, column, index, diagonal traversal size
RecursiveCantor(0, 0, 1, 0);

int n;

while( scanf("%d", &n) == 1 ){
printf("TERM %d IS %d/%d\n", n, OrderedCantor[n].numerator, OrderedCantor[n].denominator );
}

return 0;
}



### Code Recursive DP Sequence Memoize:

/***********************************************************************
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 264 - Count on Cantor
*
* Technique: Zig Zag / Spiral Diagonal traversal,
*            Recursive Dynamic Programming
***********************************************************************/

#include<stdio.h>
#include<string.h>

#define M 10000000

/**
* Table to construct the sequence
*/
struct CantorTable{
int numerator;
int denominator;
};

// Holds cantor values from 1 to M by index
CantorTable OrderedCantor[M];

void RecursiveCantor(int r, int c, int index, int diagonal){

// base case
if( r < 0 || c < 0 || index > M ){
return;
}

// change from table traversed DP
OrderedCantor[index].numerator = r + 1;
OrderedCantor[index].denominator = c + 1;

// Amount of times to travel in diagonal
++diagonal;

if(r == 0){

// when odd move down left
if(c % 2){
for(int i = 0; i < c + diagonal; ++i){
++index;
r = r + 1;
c = c - 1;
RecursiveCantor( r, c, index, diagonal );
}
}

// when even Move right
else{
++index;
c = c + 1;
RecursiveCantor( r, c, index, diagonal );
}
}

else if(c == 0){

// when odd move down
if(r % 2){
++index;
r = r + 1;
RecursiveCantor( r, c, index, diagonal );
}

// when even Move up right
else{
for(int i = 0; i < r + diagonal; ++i){
++index;
r = r - 1;
c = c + 1;
RecursiveCantor( r, c, index, diagonal );
}
}

}

}

int main() {

// Pass in row, column, index, diagonal traversal size
RecursiveCantor(0, 0, 1, 0);

int n;

while( scanf("%d", &n) == 1 ){
printf("TERM %d IS %d/%d\n", n, OrderedCantor[n].numerator, OrderedCantor[n].denominator );
}

return 0;
}


## UVA Problem 10921 – Find the Telephone Solution

UVA Problem 10921 – Find the Telephone Solution:

Solving Technique:

Given a string of alphabets (A-Z), 1, 0, and Hyphens( – ) task for this problem is to find numbers represented by a character from the given table.

1, 0, Hyphen should not be changed since they aren’t in table. Just change a character with its integer representation from table. Easiest way to do this is create a mapping array with pattern from the table. Then map the pattern to each character and print result.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

1-HOME-SWEET-HOME
MY-MISERABLE-JOB


Output:

1-4663-79338-4663
69-647372253-562


### Code:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 10921 - Find The Telephone
* Technique: Mapping character from an array to another.
*/

#include<stdio.h>
#include<string.h>

#define N 128

static char mappingArray[N] = "22233344455566677778889999";

static char table[N];
static char input[N];

int main(){

/**
* The code below fill 'A' to 'O' with appropriate values.
* Not necessary when using mapping array.
*/
/*
for(int i = 0; i <= 15; ++i ){
table[64 + i] = k;
if(i % 3 == 0) ++k;
}
*/

int k = 0;

/**
* Map each character index to integer value from mapping array.
*/
for(int i = 'A'; i <= 'Z'; ++i)
table[i] = mappingArray[k++];

/**
* These characters don't change.
*/
table['-'] = '-';
table['0'] = '0';
table['1'] = '1';

/**
* Take input, then iterate through each character
* and print the character indexed value from table.
*/
while( gets(input) ){
for(int i = 0; input[i]; ++i)
printf("%c", table[ input[i] ]);
printf("\n");
}

return 0;
}


## UVA Problem 10008 – What’s Cryptanalysis? Solution

UVA Problem 10008 – What’s Cryptanalysis? Solution:

Solving Technique:

Task for this problem is to find occurrence of each alphabetical character. Uppercase and lowercase are treated as same characters. No other characters except for upper and lower case characters must be counted.

After counting their occurrence print them in most to least order. If multiple characters have same character count then print the characters alphabetically. Ex: If E occurred 2 times, A occurred 2 times then, print A first then E.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

3
This is a test.
Count me 1 2 3 4 5.
Wow!!!! Is this question easy?


Output:

S 7
T 6
I 5
E 4
O 3
A 2
H 2
N 2
U 2
W 2
C 1
M 1
Q 1
Y 1


### Code:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 10008 - What's Cryptanalysis
* Technique: Counting character occurrence in string,
*            Lexicographical / alphabetically sort characters,
*/

#include<stdio.h>
#include<string.h>
#include<algorithm>

#define N 256

// Holds each character and its occurrence in string.
struct collection{
char character;
int occurence;
} node[26];

static char s[N];

// Sorts based on occurrence, if occurrence count is same
// then sort using lexicographical / alphabetically first character.
int cmp(collection a, collection b){

if(a.occurence < b.occurence)
return 0;
else if(a.occurence > b.occurence)
return 1;
else
return (a.character <= b.character) ? 1 : 0;

}

int main(){

// O(1)
// Represent character by its ASCII value index.
for(int i = 'A'; i <= 'Z'; ++i)
node[i].character = i;

int n;

scanf("%d", &n);
getchar();

while( n-- ){

gets( s );

// Convert string to uppercase O(n)
for(int i = 0; s[i]; ++i){
if( s[i] >= 'a' && s[i] <= 'z')
s[i] = s[i] - 32;
}

// O(n)
// Counting using character index
for(int i = 0; s[i]; ++i){
if( s[i] >= 'A' && s[i] <= 'Z' )
++node[ s[i] ].occurence;
}

}

// Sorts occurrence most to least and in lexicographic order.
std::sort(node + 'A', node + 'Z', cmp);

// O(1)
// Prints sorted occurrence most to least.
// Also prints characters in lexicographic
// order if two occurrence are same.
for(int i = 'A'; i <= 'Z'; ++i){
if(node[i].occurence)
printf("%c %d\n", node[i].character, node[i].occurence);
}

return 0;
}


## UVA Problem 11713 – Abstract Names Solution

UVA Problem 11713 – Abstract Names Solution:

Solving Technique:

For input of n there will be 2*n lines. Task is to match a pair of lines. The vowels a, e, i, o, u in the strings doesn’t matter. Except these the rest of both strings should match.

That means each character in a index from both strings should match. Their order needs to be same. Except vowels if the match then output is yes. Otherwise the output is no.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

5
pele
polo
pele
pola
ronaldo
ronaldino
pele
pelet
pele
bele


Output:

Yes
Yes
No
No
No


### Code:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 11713 - Abstract Names
* Technique: Removing specific characters from string
*            by copying into another string.
*/

#include<stdio.h>
#include<string.h>

#define N 36

static char s[N];
static char t[N];

static char tmpS[N];
static char tmpT[N];

int main(){

int n;

scanf("%d", &n);
getchar();

while( n-- ){

gets(s);
gets(t);

// If their length is not the same then output is no.
if( strlen(s) != strlen(t) ){
printf("No\n");
continue;
}

// Reset memory.
int j = 0, k = 0;
memset(tmpS, 0, sizeof tmpS);
memset(tmpT, 0, sizeof tmpT);

// Remove vowels from the first string.
for(int i = 0; s[i]; ++i){
if(s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u')
continue;
else
tmpS[j++] = s[i];
}

// Remove vowels from the second the string.
for(int i = 0; t[i]; ++i){
if(t[i] == 'a' || t[i] == 'e' || t[i] == 'i' || t[i] == 'o' || t[i] == 'u')
continue;
else
tmpT[k++] = t[i];
}

// After removing the vowels the strings should match.
if( strcmp(tmpS, tmpT) == 0 )
printf("Yes\n");
else
printf("No\n");

}

return 0;
}


## UVA Problem 10035 – Primary Arithmetic Solution

UVA Problem 10035 – Primary Arithmetic Solution:

Solving Technique:

For this problem two integers less than 10 digits are given. The task is to find how many carry operations occur when adding.

This problem may be solved using strings since input integer is less than 10 digits. It can be solved with out using array. My implementation is using character array to add and count carry operations.

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

Input:

123 456
555 555
123 594
0 0


Output:

No carry operation.
3 carry operations.
1 carry operation.


### Code:

/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 10035 - Primary Arithmetic
*            Making two strings of same length,
*/

#include<stdio.h>
#include<string.h>

#define N 128

static char s[N];
static char t[N];

static char output[N];

int main(){

while( scanf("%s", s) && scanf("%s", t) ){

// Although a major flaw input beginning with 0 and
// two strings matching. But for this problem it doesn't matter.
if( strcmp(s,t) == 0 && s[0] == '0' )
break;

int lens = strlen(s);
int lent = strlen(t);

// Shift each characters in string right by padding length.
if( lens > lent ){

int padding = lens - lent;

for(int i = lent - 1; i >= 0; --i)

for(int i = 0; i < padding; ++i)
t[i] = '0';
}

else if( lens < lent ){

int padding = lent - lens;

for(int i = lens - 1; i >= 0; --i)

for(int i = 0; i < padding; ++i)
s[i] = '0';
}

int maxlen;
if(lens > lent)
maxlen = lens;
else maxlen = lent;

int carry = 0;
int c = 0;
int sum = 0;

// Add two Strings, if a carry operation occurs
// then add that to the count.
for(int i = maxlen - 1; i >= 0; --i){

sum += s[i] - '0' + t[i] - '0';

sum = sum + carry;

carry = sum / 10;

if(carry)
++c;

sum = 0;

}

if(!c)
printf("No carry operation.\n");
else if( c > 1 )
printf("%d carry operations.\n", c);
else
printf("%d carry operation.\n", c);

}

return 0;
}