P3131 [USACO16JAN]子共七Subsequences Summing to Sevens

Posted 2020-12-01 xiongchongwen

题目描述

Farmer John‘s NN 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 61…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 NN (1 leq N leq 50,0001≤N≤50,000). The next NN

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

0 ldots 1,000,0000…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.

输入输出样例

输入 #1复制

7
3
5
1
6
2
14
10

输出 #1复制

5

说明/提示

In this example, 5+1+6+2+14 = 28.

 

 

#include<iostream>
#include<cstdio>
using namespace std;
int s,n,i,f[15],a,ans;
int main(){
	cin>>n;
	for(int k=1;k<=15;k++){
	    f[k]=-1;
	}
	f[0]=0;
	for(i=1;i<=n;i++){
		cin>>a;
		s=(s+a)%7;
		if(f[s]==-1){f[s]=i;}
		else{ans=max(ans,i-f[s]);}
	}
	cout<<ans;
}