题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4648

Magic Pen 6

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Problem Description

In HIT, many people have a magic pen. Lilu0355 has a magic pen, darkgt has a magic pen, discover has a magic pen. Recently, Timer also got a magic pen from seniors.

At the end of this term, teacher gives Timer a job to deliver the list of N students who fail the course to dean‘s office. Most of these students are Timer‘s friends, and Timer doesn‘t want to see them fail the course. So, Timer decides to use his magic pen
to scratch out consecutive names as much as possible. However, teacher has already calculated the sum of all students‘ scores module M. Then in order not to let the teacher find anything strange, Timer should keep the sum of the rest of students‘ scores module
M the same.

Plans can never keep pace with changes, Timer is too busy to do this job. Therefore, he turns to you. He needs you to program to "save" these students as much as possible.

Input

There are multiple test cases.

The first line of each case contains two integer N and M, (0< N <= 100000, 0 < M < 10000),then followed by a line consists of N integers a₁,a₂,...a_n (-100000000 <= a₁,a₂,...a_n <= 100000000) denoting
the score of each student.(Strange score? Yes, in great HIT, everything is possible)

Output

For each test case, output the largest number of students you can scratch out.

Sample Input

2 3
1 6
3 3
2 3 6
2 5
1 3

Sample Output

1
2
0

Hint

The magic pen can be used only once to scratch out consecutive students.

Source

2013 Multi-University Training Contest 5

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简单题，len从大到小，(sum[j+len]-sum[j])%M==0 len即为所求。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAX=100010;
int sum[MAX];
int main()
{
    int N,M;
    while(cin>>N>>M)
    {
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=N;i++)
        {
            int x;
            scanf("%d",&x);
            sum[i]=sum[i-1]+x;
        }
        bool flag=true;
        for(int i=N;i>=0&&flag;i--)
        {
            for(int j=0;j+i<=N;j++)
            {
                if((sum[j+i]-sum[j])%M==0)
                {
                    cout<<i<<endl;
                    flag=false;
                    break;
                }
            }
        }
    }
    return 0;
}

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