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

样例输出

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+;

const int dis[][]={{,},{-,},{,-},{,}};

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m;

ll a[maxn],pre[],last[],now;

int main()

{

    int i,j,k,t;

    scanf("%d",&n);

    rep(i,,n){

        scanf("%lld",&a[i]);

        now=(now+a[i])%;

        if(pre[now])last[now]=i;

        else pre[now]=i;

    }

    ll ma=;

    rep(i,,)

    {

        if(pre[i]&&last[i])ma=max(ma,last[i]-pre[i]);

    }

    printf("%lld\n",ma);

    //system ("pause");

    return ;

}

巴特西

Subsequences Summing to Sevens