Code
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main() {
int T;
cin >> T;
while (T--) {
ll n , k , x , y ;
cin >> n >> k ;
if (k >= n) {
// We can increment all elements to at least 1
x = k - n; // Remaining operations after making all elements >= 1
y = 1; // The AND of all elements will be 1
cout << x * y << endl;
}
else {
// We cannot make all elements >= 1, so Y will be 0
cout << 0 << endl;
}
}
return 0;
}
Information
- Submit By
- Type
- Pretest
- Problem
- P1092 Bitwise AND
- Language
- C++20 (G++ 13.2.0)
- Submit At
- 2024-09-05 16:14:57
- Judged At
- 2024-11-11 03:01:15
- Judged By
- Score
- 10
- Total Time
- 1ms
- Peak Memory
- 336.0 KiB