[USACO16JAN] Subsequences Summing to Sevens S

题目描述

Farmer John's $N$ cows are standing in a row, as they have a tendency to do from time to time. Each cow is labeled with a distinct integer ID number so FJ can tell them apart. FJ would like to take a photo of a contiguous group of cows but, due to a traumatic childhood incident involving the numbers $1 \ldots 6$, he only wants to take a picture of a group of cows if their IDs add up to a multiple of 7.

Please help FJ determine the size of the largest group he can photograph.

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。

输入格式

The first line of input contains $N$ ($1 \leq N \leq 50,000$). The next $N$

lines each contain the $N$ integer IDs of the cows (all are in the range

$0 \ldots 1,000,000$).

输出格式

Please output the number of cows in the largest consecutive group whose IDs sum

to a multiple of 7. If no such group exists, output 0.

样例 #1

样例输入 #1

7
3
5
1
6
2
14
10

样例输出 #1

5

提示

In this example, 5+1+6+2+14 = 28. #include<bits/stdc++.h> using namespace std; const int N=1e5+1; long long int n,mx,a[N],sum[N];

signed main(){ cin>>n; for(int i=1;i<=n;i++){ cin>>a[i]; sum[i]=sum[i-1]+a[i]; }

for(int l=1;l<=n;l++){
	for(int r=l+mx;r<=n;r++){
		if((sum[r]-sum[l-1])%7==0)
		mx=r-l+1;
	}
}
cout<<mx;	
return 0;

}