# HackerEarth Valid Chess Board problem solution

In this HackerEarth Valid Chess Board problem solution, you are given a tiled chart paper of N rows and M columns. There are a total of N x M tiles in it. Each tile is colored either black or white. Now, you need to count how many ways are there to cut valid chessboards of size 8 x 8 out of this chart paper.

## HackerEarth Valid Chess Board problem solution.

`#include<bits/stdc++.h>#define LL long long int#define M 1000000007#define endl "\n"#define eps 0.00000001LL pow(LL a,LL b,LL m){ a%=m;LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}using namespace std;char a[1001][1001];int f(int x1,int y1,int x2,int y2)  {    for(int i = x1; i <= x2; i++)   {        for(int j = y1 + 1; j <= y2; j++)   {            if(a[i][j] == a[i][j - 1])  {                return 0;            }        }    }    for(int j = y1; j <= y2; j++)   {        for(int i = x1 + 1; i <= x2; i++)   {            if(a[i][j] == a[i - 1][j])  {                return 0;            }        }    }    return 1;}int main()  {    ios_base::sync_with_stdio(0);    cin.tie(0);    int n, m;    cin >> n >> m;    for(int i = 0; i < n; i++)  {        for(int j = 0; j < m; j++)  {            cin >> a[i][j];        }    }    int ans = 0;    for(int i = 0; i < n; i++)  {        for(int j = 0; j < m; j++)  {            if(i + 8 <= n && j + 8 <= m) {                ans = ans + f(i , j , i + 7 , j + 7);            }        }    }    cout << ans;}`

### Second solution

`#include <bits/stdc++.h>using namespace std;const int N = 1E3 + 5;string mat[N];bool f(int x, int y, int n, int m) {    for(int i = x; i < x + 8; i ++) {        for(int j = y; j < y + 7; j ++) {            if(mat[i][j] == mat[i][j + 1] || (i < x + 7 && mat[i][j] == mat[i + 1][j]))                return 0;        }    }    return 1;}int main() {    ios_base::sync_with_stdio(false);    cin.tie(NULL);    int n, m;    cin >> n >> m;    for(int i = 0; i < n; i ++)        cin >> mat[i];    int cnt = 0;    for(int i = 0; i < n; i ++) {        for(int j = 0; j < m; j ++) {            if(i + 8 <= n && j + 8 <= m && f(i, j, n, m))                cnt ++;        }    }    cout << cnt;    return 0;}`