题目：P3131 [USACO16JAN]Subsequences Summing to Sevens S

题目：[USACO16JAN]Subsequences Summing to Sevens S - 洛谷

题目大意

给定一个序列，要求计算出能被7整除的最长序列

坑点

无

数据范围

int足以

思路

可以运用前缀和，再进行判断

代码

#include<iostream>
using namespace std;
long long int a[50010];
long long int sum[50010];
int main()
{int n;cin>>n;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=1;i<=n;i++){sum[i]=sum[i-1]+a[i];//前缀和 }for(int i=1;i<=n;i++){sum[i]=sum[i]%7;//取余 }int maxn=-1;//给定一个数 for(int i=0;i<=6;i++){int l=0,r=0;//定义左右端点 if(i==0){for(int j=n;j>=1;j--){if(sum[j]==i)//判断 {r=j;break; }}maxn=max(maxn,r);//最大值 continue;}for(int j=1;j<=n;j++){if(sum[j]==i){l=j;break; }}for(int j=n;j>=1;j--){if(sum[j]==i){r=j;break; }}if(l!=0&&r!=0){maxn=max(maxn,r-l);}}if(maxn==-1){cout<<0;}else{cout<<maxn;}
}

总结

运用前缀和运算，再对余数判断

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