## UVA Problem 424 – Integer Inquiry Solution

UVA Problem 424 – Integer Inquiry Solution:

Solving Technique:

This problem requires adding integers. The integers are very long, meaning they won’t fit even in long long. So the addition needs to be done using arrays.

It can be solved using integers arrays but my solution uses character array. Code is explained in the comments.

Similar to this problem UVA 10035 Primary Arithmetic and UVA 713 – Adding Reversed Numbers.

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:

```123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0
```

Output:

```370370367037037036703703703670
```

### Code:

```/**
* Author:    Asif Ahmed
* Site:      https://quickgrid.wordpress.com
* Problem:   UVA 424 - Integer Inquiry
* Technique: Adding Multi String Integer characters column wise
*/

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

#define N 160

static char s[N][N];
static char output[N];

int main(){

int i = 0;

int maxlen = 0;

// Finish taking all the inputs.
while( gets(s[i]) ){

// Exit for input of 0.
if( s[i] == '0' && ! s[i] )
break;

int len = strlen( s[i] );

// Get the max length to create padding.
if( len > maxlen )
maxlen = len;

++i;
}

// Save rows
int rows = i;

// Create padding for each of the strings.
for(int j = 0; j < i; ++j){

int temp = strlen( s[j] );

if( temp != maxlen ){

// Shift by this many spaces to create 0's in front.
int padding = maxlen - temp;

for(int k = temp - 1; k >= 0; --k)

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

int carry = 0, z = 0;

for(int j = maxlen - 1; j >= 0; --j){

int sum = 0;

for(int i = 0; i < rows; ++i)
sum += s[i][j] - '0';

// Add if any previous  carry
sum = sum + carry;

output[z++] = sum % 10 + '0';

// get the carry for adding to next column
carry = sum / 10;

}

// Print if any carry first
if(carry)
printf("%d", carry);

// Then print what ever character is left
for(int i = z - 1; i >= 0; --i){
if(output[i] >= '0' && output[i] <= '9')
printf("%c", output[i]);
}
printf("\n");

return 0;
}
```