HackerEarth Flip the World problem solution

YASH PAL,
In this HackerEarth Flip, the world problem-solution Flip the world is a game. In this game, a matrix of size N * M is given, which consists of numbers. Each number can be 1 or 0 only. The rows are numbered from 1 to N, and the columns are numbered from 1 to M.
The following steps can be called as a single move.
Select two integers x,y (1 <= x <= N and 1 <= y <= M) i.e. one square on the matrix.
All the integers in the rectangle denoted by (1,1) and (x,y) i.e. rectangle having top-left and bottom-right points as (1,1) and (x,y) are toggled(1 is made 0 and 0 is made 1).
For example, in this matrix (N = 4 and M = 3)
101
110
101
000
if we choose x = 3 and y = 2, the new state of matrix would be
011
000
011
000
For a given state of matrix, aim of the game is to reduce the matrix to a state where all numbers are 1. What is minimum number of moves required.
HackerEarth Flip the World problem solution

HackerEarth Flip the World problem solution.

#include <bits/stdc++.h>

using namespace std;

char arr [20 + 1][30 + 1];

int main()
{
    int t , n , m , ans = 0;

    cin >> t;

    while(t--){

        ans = 0;
        cin >> n >> m;

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

                cin >> arr[i][j];
            }
        }

        for(int i = n - 1; i >= 0; i--){
            for(int j = m - 1; j >= 0; j--){

                if(arr[i][j] == '0'){   
                    ans++;            
                    for(int k = 0; k <= i; k++){ 
                        for(int h = 0; h <= j; h++){

                            if(arr[k][h] == '1') arr[k][h] = '0';
                            else arr[k][h] = '1';
                        }
                    }
                }
            }
        }

        cout << ans << "n";
    }

    return 0;
}

Second solution

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<vector>
#include<cassert>
#include<sstream>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define clr(x) x.clear()
#define sz(x) ((int)(x).size())
#define F first
#define S second
#define REP(i,a,b) for(i=a;i<b;i++)
#define rep(i,b) for(i=0;i<b;i++)
#define rep1(i,b) for(i=1;i<=b;i++)
#define pdn(n) printf("%dn",n)
#define sl(n) scanf("%lld",&n)
#define sd(n) scanf("%d",&n)
#define pn printf("n")
typedef pair<int,int> PII;
typedef vector<PII> VPII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef long long LL;
#define MOD 1000000007
LL mpow(LL a, LL n) 
{LL ret=1;LL b=a;while(n) {if(n&1) 
    ret=(ret*b)%MOD;b=(b*b)%MOD;n>>=1;}
return (LL)ret;}
int main()
{
    int t;
    sd(t);
    while(t--)
    {
        int n,m,i,j,a[60][60]={};
        vector <string> board;
        sd(n),sd(m);
        board.resize(n);
        for(i=0; i<n; i++)
            cin >> board[i];
        for(i=n-1;i>=0;--i) 
            for(j=m-1;j>=0;--j) 
            { 
                if(i==n-1&&j==m-1) 
                    a[i][j]=0; 
                else if(i==n-1) 
                    a[i][j]=a[i][j+1]; 
                else if(j==m-1) 
                    a[i][j]=a[i+1][j]; 
                else 
                    a[i][j]=a[i+1][j]+a[i][j+1]-a[i+1][j+1]; 
                if((a[i][j]%2==1&&board[i][j]=='1')||(a[i][j]%2==0&&board[i][j]=='0')) 
                    a[i][j]++; 
            } 
        pdn(a[0][0]);
    }
    return 0;
}
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