Modulo Sum(背包 + STL)

 Modulo Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence of numbers a₁, a₂, ..., a_n, and a number m.

Check if it is possible to choose a non-empty subsequence a_ij such that the sum of numbers in this subsequence is divisible bym.

Input

The first line contains two numbers, n and m (1 ≤ n ≤ 10⁶, 2 ≤ m ≤ 10³) — the size of the original sequence and the number such that sum should be divisible by it.

The second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁹).

Output

In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.

Examples

input

3 5 1 2 3

output

YES

input

1 6 5

output

NO

input

4 6 3 1 1 3

output

YES

input

6 6 5 5 5 5 5 5

output

YES

Note

In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.

In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.

In the third sample test you need to choose two numbers 3 on the ends.

In the fourth sample test you can take the whole subsequence.

题意：

取任意个数的和能否组成M的倍数；宇神用背包写的，参谋了下，很聪明的解法，还可以用set做；

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 1e3 + ;
int num[MAXN * MAXN];
int v[MAXN * MAXN];
int dp[ * MAXN][MAXN];
int val[MAXN * MAXN];
int main(){
    int n, m;
    while(~scanf("%d%d", &n, &m)){
        memset(num, , sizeof(num));
        memset(v, , sizeof(v));
        memset(dp, , sizeof(dp));
        memset(val, , sizeof(val));
        int tp = ;
        int temp;
        int ans = ;
        for(int i = ; i < n; i++){
            scanf("%d", &temp);
            if(temp == m){
                ans = ;
            }
            if(!num[temp % m])
                v[tp++] = temp % m;
            num[temp % m]++;
        }
        n = tp;
        /*
        printf("tp = %d\n",tp);
        for(int i = 0; i < tp; i++){
            printf("v = %d num = %d \n", v[i], num[v[i] ]);
        }puts("");
        */
        tp = ;
        for(int i = ; i < n; i++){
            for(int j = ; j <= num[v[i]]; j <<= ){
                if(j * v[i] % m == ){
                    ans = ;
                }
                num[v[i]] -= j;
                val[tp++] = j * v[i] % m;
            }
            if(num[v[i]] > ){
                if(v[i] * num[v[i]] % m == )
                    ans = ;
                val[tp++] = v[i] * num[v[i]] % m;
            }
        }
        /*
        printf("tp = %d\n",tp);
        for(int i = 0; i < tp; i++){
            printf("v = %d num = %d \n", val[i], num[v[i] ]);
        }puts("");
        */
        dp[][v[]] = ;dp[][] = ;
    //    printf("ans = %d\n",ans);
        for(int i = ; i < tp - ; i++){
            for(int j = ; j <= m; j++){
                if(dp[i][j]){
                //    printf("i = %d j = %d\n",i,j);
                    dp[i + ][j] = ;
                    if((j + val[i + ]) % m == ){
                        ans = ;
                    }
                    dp[i + ][(j + val[i + ]) % m] = ;
                }
            }
        }
        if(ans)puts("YES");
        else puts("NO");
    }
    return ;
}

set:

#include<iostream>
#include<cmath>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
using namespace std;
const int MAXN = 1e6 + ;
int main(){
    int n, m;
    while(~scanf("%d%d", &n, &m)){
        int temp, ans = ;
        set<int>st, _st;
        st.insert();
        set<int>::iterator iter;
        for(int i = ; i < n; i++){
            scanf("%d", &temp);
            for(iter = st.begin(); iter != st.end(); iter++){
                if((*iter + temp) % m == ){
                    printf("YES\n");
                    return ;
                }
                _st.insert((*iter + temp) % m);
            }
            st.insert(_st.begin(), _st.end());
            _st.clear();
        }
        puts("NO");
    }
    return ;
}

vector:

#include<iostream>
#include<cmath>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<vector>
using namespace std;
const int MAXN = 1e6 + 100;
int vis[MAXN];
int main(){
	int n, m;
	while(~scanf("%d%d", &n, &m)){
		int temp, ans = 0;
		vector<int>st, _st;
		st.push_back(0);
		memset(vis, 0, sizeof(vis));
		for(int i = 0; i < n; i++){
			scanf("%d", &temp);
			if(ans)continue;
			for(int i = 0; i < st.size(); i++){
				int x = (st[i] + temp) % m;
				if(x == 0){
					ans = 1;
					break;
				}
				if(!vis[x])vis[x] = 1,_st.push_back(x);
			}
			for(int i = 0; i < _st.size(); i++){
				st.push_back(_st[i]);
			}
			_st.clear();
		}
		if(ans)puts("YES");
		else puts("NO");
	}
	return 0;
}