COJ: A-SUM（贪心）

sum
Time limit :1000 ms , Memory limit :131072 kB

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`.

Simple Input

2
3 3
1 2 3
5 7
6 6 6 6 6

Simple Output

YES
NO

Hint
CSU-ACM2017暑期集训热身训练1
原题链接：A-sum

思路：遍历加贪心，查找序列中是否存在一个子序列能被m整除。
但此方法的时间复杂度是O(n!)，对于较长的序列可能会超时^_^

Ac代码

#include<iostream>
using namespace std;
#define maxn 100005
int x[maxn]={0};

int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--)
    {
        int n, m, sum, flag=0;
        cin >> n >> m;
        for(int i=0; i<n; i++)
        {
            cin >> x[i];
            if(x[i] % m ==0) flag=1;
        }
        if(flag==1) cout << "YES\n";
        else{
            for(int i=0; i<n; i++){
                sum=0;
                if(flag==1) break;
                for(int j=i; j<n; j++)
                {
                    sum+=x[j];
                    if(sum % m==0) {flag=1; break;}
                }
            }
            if(flag==1) cout << "YES\n";
            else cout << "NO\n";
        }
    }
    return 0;
}