P3131 [USACO16JAN]Subsequences Summing to Sevens S-二分+前缀和

给你n个数，分别是a[1],a[2],…,a[n]。求一个最长的区间[x,y]，使得区间中的数(a[x],a[x+1],a[x+2],…,a[y-1],a[y])的和能被7整除。输出区间长度。若没有符合要求的区间，输出0。

思路：原本想从长度为n,n-1,n-2……2,1 这样暴力搜索一下，但是最坏情况是O(n^2)，果然超时了
于是可以考虑用二分，见代码

#include <bits/stdc++.h>
using namespace std;
#define de(x) cout<<x<<" ";
#define Pu puts("");
#define sf(x) scanf("%d",&x);
typedef long long ll;
const int N=1e6+10,mod=100003;
const int inf=0x3f3f3f3f;
int a[N],pre[N];
int n,m,ans;
int fun(int len){int t;for(int i=len;i<=n;i++){t=(pre[i]-pre[i-len]+7)%7;if(t==0) return 1;}return 0;
}
int main(){cin>>n;int x;for(int i=1;i<=n;i++){sf(x)a[i]=x%7;pre[i]=(pre[i-1]+a[i])%7;}int l=0,r=n+1,mid;while(l<r){mid=(l+r)>>1;if(fun(mid)) l=mid+1;else r=mid;}printf("%d\n",l-1);return 0;
}

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