2 solutions
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0
#include<bits/stdc++.h> using namespace std; int n, x, a[1000010],sum=0; int main(){ cin >> n; for(int i = 0;i < n; i++){ cin >> a[i]; } cin >> x; sort(a, a + n); for(int i = 0;i < n; i++){ int idx = lower_bound(a, a + n, x - a[i]) - a; if(a[idx] == x - a[i]) sum++; } cout << sum / 2; return 0; }
Information
- ID
- 1102
- Time
- 1000ms
- Memory
- 20MiB
- Difficulty
- 6
- Tags
- # Submissions
- 76
- Accepted
- 21
- Uploaded By