1 solutions
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1
#include<bits/stdc++.h> using namespace std; const int N=1e5+10; int a[N],b[N],c[N],cnt[N]; int main() { int n; cin>>n; for(int i=1;i<=n;i++) { cin>>a[i]; } for(int i=1;i<=n;i++) { cin>>b[i]; } for(int i=1;i<=n;i++) { cin>>c[i]; cnt[b[c[i]]]++;//对应的b[c[i]]可以使用的次数增加一个 } long long ans=0; for(int i=1;i<=n;i++) { ans=ans+cnt[a[i]]; //和a[i]匹配的b[c[i]]的数目 } cout<<ans; return 0; }
- 1
Information
- ID
- 2372
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 3
- Tags
- (None)
- # Submissions
- 9
- Accepted
- 5
- Uploaded By