1 solutions
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1
#include<bits/stdc++.h> using namespace std; const int N=1e5+10; int a[N]; set<int> h; int main() { int n; cin>>n; for(int i=1;i<=n;i++) cin>>a[i]; int x; cin>>x; int cnt=0; for(int i=1;i<=n;i++){ if(h.count(x-a[i])) //统计之前的差值有没有出现过 cnt++; h.insert(a[i]); //插入当前值 } cout<<cnt; return 0; }
- 1
Information
- ID
- 159
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 1
- Tags
- (None)
- # Submissions
- 20
- Accepted
- 8
- Uploaded By