hdu-5776 sum(同余)

题目链接：

sum

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 131072/131072 K (Java/Others)

Problem Description

Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO

 

Input

The first line of the input has an integer T (1≤T≤10), which represents the number of test cases. 
For each test case, there are two lines:
1.The first line contains two positive integers n, m (1≤n≤100000, 1≤m≤5000).
2.The second line contains n positive integers x (1≤x≤100) according to the sequence.

 

Output

Output T lines, each line print a YES or NO.

 

Sample Input

 

2

3 3

1 2 3

5 7

6 6 6 6 6

 

Sample Output

 

YES

NO

 

题意：

 

问是否存在字串的和为m倍数;

 

思路：

 

sum[r]-sum[l]=k*m;sum[r]=sum[l](modm);

前缀和同余就存在;

 

AC代码：

 

/************************************************
┆  ┏┓　　　┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　┻　　　┃ ┆
┆┗━┓　  　┏━┛ ┆
┆　　┃　  　┃　　┆　　　　　　
┆　　┃　  　┗━━━┓ ┆
┆　　┃　 AC代马 　　┣┓┆
┆　　┃　        　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */  

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=(1<<8);
const double eps=1e-8;

int vis[5600];

int main()
{       

        int t;
        read(t);
        while(t--)
        {
            mst(vis,0);
            int n,m,sum=0,flag=0,x;
            read(n);read(m);
            vis[0]=1;
            For(i,1,n)
            {
                read(x);
                sum+=x;
                int temp=sum%m;
                if(vis[temp])flag=1;
                vis[temp]=1;
            }
            if(flag)cout<<"YES\n";
            else cout<<"NO\n";
        }

        return 0;
}