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You do not need to have any conditions to complete this problem.
int main() { int n; int firstDiagonal=0; int secondDiagonal=0; cin >> n; int count=n-1; vector< vector<int> > a(n,vector<int>(n)); for(int a_i = 0;a_i < n;a_i++){ for(int a_j = 0;a_j < n;a_j++){ cin >> a[a_i][a_j]; } } for(int i=0; i<n; i++) { firstDiagonal+=a.at(i).at(i); secondDiagonal+=a.at(count).at(i); count--; } cout << abs(firstDiagnol-secondDiagnol); return 0; }
Because the diagonal are always going to be positions we know when we have the size of the matrix.
Diagonal Difference
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You do not need to have any conditions to complete this problem.
Because the diagonal are always going to be positions we know when we have the size of the matrix.