CodeForce 577B Modulo Sum
You are given a sequence of numbers a1, a2, ..., an, and a number m.
Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.
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, ..., an (0 ≤ ai ≤ 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
3 5
1 2 3
YES
1 6
5
NO
4 6
3 1 1 3
YES
6 6
5 5 5 5 5 5
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 subsequen
OJ-ID:
CodeForce 577B
author:
Caution_X
date of submission:
20191019
tags:
dp
description modelling:
给一个序列,找一个子序列使之mod m =0
major steps to solve it:
dp[i]:表示取模后得到了i
1.遍历序列,对每一个元素更新dp,判断能否得到dp[0]
AC code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[],dp[],tmp[];
int main()
{
ll n,m;
scanf("%lld%lld",&n,&m);
for(int i=;i<n;i++) {
scanf("%lld",&a[i]);
}
for(int i=;i<n;i++) {
if(dp[]) break;
for(int j=;j<=m-;j++) {
if(dp[j]) {
tmp[(j+a[i])%m]=;
}
}
tmp[a[i]%m]=;
for(int j=;j<=m-;j++) {
dp[j]=tmp[j];
}
}
if(dp[]) printf("YES\n");
else printf("NO\n");
return ;
}
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