**UVA Problem 417 – Word Index Solution:**

Click here to go to this problem in uva Online Judge.

**Solving Technique:**

This one of the worst code I have ever written. It is both slow and wastes huge amount of memory, but still works. I am sharing this code as an example 5 dimensional re-sizable (vector) array. There are far far far better solution than this, so I won’t explain it.

If the input is thought of as string, then for each position the character right to current character is at least current character plus one. Since there are 5 length classes 1,2,3,4,5. So for each of them I applied this logic.

** 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:**

26 1 0 83681

**Output:**

z a cat vwxyz

### Code:

/** * Author: Asif Ahmed * site: quickgrid.wordpress.com * Problem: UVA 417 - word index * Technique: Very Slow and worst possible solution. * 5 (five) dimensioanl vector integer array. 1, 2 * 3, 4 (four) dimensional integer array. */ #include <vector> #include <iostream> #include<cstdio> #include<cstring> using namespace std; #define N 26 int main() { //freopen("input.txt", "r", stdin); //freopen("output.txt", "w", stdout); vector<vector<vector<vector<vector<int> > > > > array5d; array5d.resize(N); for (int i = 0; i < N; ++i) { array5d[i].resize(N); for (int j = 0; j < N; ++j){ array5d[i][j].resize(N); for (int k = 0; k < N; ++k){ array5d[i][j][k].resize(N); for (int m = 0; m < N; ++m){ array5d[i][j][k][m].resize(N); } } } } int s1[N]; int s2[N][N]; int s3[N][N][N]; int s4[N][N][N][N]; int n = 0; for(int i = 0; i < N; ++i) s1[i] = ++n; for(int i = 0; i < N; ++i){ for(int j = i + 1; j < N; ++j){ s2[i][j] = ++n; } } for(int i = 0; i < N; ++i){ for(int j = i + 1; j < N; ++j){ for(int k = j + 1; k < N; ++k){ s3[i][j][k] = ++n; } } } for(int i = 0; i < N; ++i){ for(int j = i + 1; j < N; ++j){ for(int k = j + 1; k < N; ++k){ for(int m = k + 1; m < N; ++m){ s4[i][j][k][m] = ++n; } } } } for(int i = 0; i < N; ++i){ for(int j = i + 1; j < N; ++j){ for(int k = j + 1; k < N; ++k){ for(int m = k + 1; m < N; ++m){ for(int t = m + 1; t < N; ++t){ array5d[i][j][k][m][t] = ++n; } } } } } char input[N]; while( gets(input) ){ int len = strlen(input); switch(len){ case 1: printf("%d\n", s1[ input[0] - 'a' ] ); break; case 2: printf("%d\n", s2[ input[0] - 'a' ][ input[1] - 'a' ] ); break; case 3: printf("%d\n", s3[ input[0] - 'a' ][ input[1] - 'a' ][ input[2] - 'a' ] ); break; case 4: printf("%d\n", s4[ input[0] - 'a' ][ input[1] - 'a' ][ input[2] - 'a' ][ input[3] - 'a' ] ); break; case 5: printf("%d\n", array5d[ input[0] - 'a' ][ input[1] - 'a' ][ input[2] - 'a' ][ input[3] - 'a' ][ input[4] - 'a' ] ); break; } } return 0; }