1151B - Dima and a Bad XOR - CodeForces Solution

bitmasks brute force constructive algorithms dp *1600

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Python Code: 

n, m = list(map(int, input().split()))
T = [[] for i in range(n)]
x = 0
d = [1] * n
for i in range(n):
    a = list(map(int, input().split()))
    T[i] = a
    x ^= a[0]
 
for i in range(n):
    for j in range(m):
        if x ^ T[i][0] ^ T[i][j] != 0:
            print('TAK')
            d[i] = j + 1
            print(*d)
            exit()
 
print('NIE')

C++ Code: 

#include<bits/stdc++.h>
using namespace std;
long long n, m, i, j, a[507][507], res, row, col;
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    
    cin >> n >> m;
    
    for(i = 1; i <= n; i++)
    {
        for(j = 1; j <= m; j++)
        {
            cin >> a[i][j];
        }
        
        res ^= a[i][1];
    }
    
    if(res)
    {
        cout << "TAK\n";
        for(i = 1; i <= n; i++) cout << "1 ";
        
        return 0;
    }
    
    for(i = 1; i <= n; i++)
    {
        for(j = 2; j <= m; j++)
        {
            if(a[i][j] != a[i][1])
            {
                res = 1;
                row = i;
                col = j;
                break;
            }
        }
        
        if(res == 1) break;
    }
    
    if(res)
    {
        cout << "TAK\n";
        for(i = 1; i <= n; i++)
        {
            if(row != i) cout << "1 ";
            else cout << col << " ";
        }
        
        return 0;
    }
    
    cout << "NIE";
}


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