[USACO16JAN]子共七Subsequences Summing to Sevens

题目描述

Farmer John's NNN 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…61 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，求一个最长的区间，使得区间和能被7整除

输入输出格式

输入格式：

The first line of input contains NNN (1≤N≤50,0001 leq N leq 50,0001≤N≤50,000 ). The next NNN

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

0…1,000,0000 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

输入样例#1： 复制

7
3
5
1
6
2
14
10

输出样例#1： 复制

5

说明

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

同余的性质

(a mod m) + (b mod m) ≡ a + b (mod m)
(a mod m) - (b mod m) ≡ a - b (mod m)
(a mod m) * (b mod m) ≡ a * b (mod m)

给出序列，求出最长的区间，满足区间和为7的倍数
预处理出前缀和
(pre[j] - pre[i ]) mod 7 = 0
=> pre[i] ≡ pre[j] (mod 7)
维护7个值，分别表示模7余0到6的最前d的前缀和
扫一遍过去就行了

//2018年2月14日14:15:25
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 100001;
int n, a[N], sum[N], last[N], first[N], ans;

int main(){
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        scanf("%d", &a[i]);
        sum[i] = (sum[i-1] + a[i]) % 7;
    }
    for(int i=1;i<=n;i++)
        last[sum[i]] = i;
    for(int i=n;i>=1;i--)
        first[sum[i]] = i;
    for(int i=0;i<7;i++)
        ans = max(ans, last[i]-first[i]);
    printf("%d
", ans);

    return 0;
}