HackerEarth Bob’s confusion problem solution

YASH PAL, 31 July 202415 February 2026

In this HackerEarth Bob’s confusion problem solution You are given a 3 x 3 grid. Each number in the grid is represented by Cij where (i,j) denotes the square at the ith row from the top and jth column from the left.

Bob gets confused. According to him, there are six integers a1, a2, a3, b1, b2, b3 whose values are fixed and the number written in the square (i,j) is equal to ai + bj. You are required to determine whether Bob is correct or not.

HackerEarth Bob's confusion problem solution

HackerEarth Bob’s confusion problem solution.

#include<bits/stdc++.h>
using namespace std;

#define     F           first
#define     S           second
#define     pb          push_back
#define     lb          lower_bound
#define     ub          upper_bound
#define     vi          vector<int>
#define     all(x)      x.begin(),x.end()
#define     fix         fixed<<setprecision(10)
#define     rep(i,a,b)  for(int i=int(a);i<=int(b);i++)
#define     repb(i,b,a) for(int i=int(b);i>=int(a);i--)
#define     FastIO      ios_base::sync_with_stdio(0),cin.tie(0)

typedef double db;
typedef long long ll;

const int N=2e5+5;
const int mod=1e9+7;

void solve(){
    int a[5],b[5],c[5][5];
    rep(i,1,3) rep(j,1,3) cin>>c[i][j];
    a[1]=0;
    b[1]=c[1][1];
    b[2]=c[1][2];
    b[3]=c[1][3];
    a[2]=c[2][2]-b[2];
    a[3]=c[3][3]-b[3];
    rep(i,1,3) rep(j,1,3){
        if(a[i]+b[j]!=c[i][j]){
            cout<<"NOn";
            return;
        }
    }
    cout<<"YESn";
}

signed main(){
    FastIO;
    int t;
    cin>>t;
    while(t--) solve();
    return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 1e5 + 14;
int n, c[3][3];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        for (auto &i : c)
            for (int &j : i)
                cin >> j;
        bool ok = true;
        for (int i = 1; i < 3; ++i)
            for (int j = 1; j < 3; ++j)
                ok &= c[i][j] == c[0][j] + c[i][0] - c[0][0];
        cout << (ok ? "YESn" : "NOn");
    }
}

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