1 solutions
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0
C++ :
#include <bits/stdc++.h> using namespace std; int s[100005][2],sum[100005][2],c[100005],x[100005];//定义二维数组方便分组 int n,m,ans; int main(){ scanf ("%d %d",&n,&m);//其实m没有什么用处 for (int i=1;i<=n;i++){ scanf ("%d",&x[i]); } for (int i=1;i<=n;i++){ scanf ("%d",&c[i]); s[c[i]][i%2]++;//求出这个分组中有多少个数 sum[c[i]][i%2]=(sum[c[i]][i%2]+x[i])%10007;//事先求出累加和,注意,要mod10007 } for (int i=1;i<=n;i++){ ans=(ans+i*((s[c[i]][i%2]-2)*x[i]%10007+sum[c[i]][i%2]))%10007;//依次然后代入,注意,也要mod10007 } printf ("%d\n",ans);//最后输出 return 0; }
- 1
Information
- ID
- 10114
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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