Subsequences Summing to Sevens

Subsequences Summing to Sevens

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…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.

输入

The first line of input contains N (1≤N≤50,000). The next N lines each contain the N integer IDs of the cows (all are in the range 0…1,000,000).

输出

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.

You may want to note that the sum of the IDs of a large group of cows might be too large to fit into a standard 32-bit integer. If you are summing up large groups of IDs, you may therefore want to use a larger integer data type, like a 64-bit "long long" in C/C++.

样例输入

7
3
5
1
6
2
14
10

样例输出

5

提示

In this example, 5+1+6+2+14 = 28. 

分析：(i+j)%k=i%k,则j是k的倍数；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <bitset>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e6+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m;
ll a[maxn],pre[7],last[7],now;
int main()
{
    int i,j,k,t;
    scanf("%d",&n);
    rep(i,1,n){
        scanf("%lld",&a[i]);
        now=(now+a[i])%7;
        if(pre[now])last[now]=i;
        else pre[now]=i;
    }
    ll ma=0;
    rep(i,0,6)
    {
        if(pre[i]&&last[i])ma=max(ma,last[i]-pre[i]);
    }
    printf("%lld\n",ma);
    //system ("pause");
    return 0;
}