Codeforces Codeforces Round #319 (Div. 2) B. Modulo Sum 背包dp-白红宇

Codeforces Codeforces Round #319 (Div. 2) B. Modulo Sum 背包dp

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发布时间：2019-06-30

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B. Modulo Sum

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/577/problem/B

Description

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.

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, ..., 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.

Sample Input

3 5

1 2 3

Sample Output

YES

HINT

题意

给你一堆数，然后问你是否有一些数加起来%m==0

题解：

当成背包dp做，空间为m，每一个物品的代价为a[i]就好了

注意滚动数组的时候，不要转移的时候被自己的状态转移了，注意一下就好了

代码：

//qscqesze#pragma comment(linker, "/STACK:1024000000,1024000000")#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  #include 
                  #include 
                   
                    typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 1000006#define mod 1001#define eps 1e-9#define pi 3.1415926int Num;//const int inf=0x7fffffff;const ll inf=999999999;inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}//*************************************************************************************int a[maxn];bool dp[maxn][2];int main(){ int n=read(),m=read(); for(int i=1;i<=n;i++) a[i]=read(),a[i]%=m; int flag = 0; for(int i=1;i<=n;i++) { dp[a[i]][1-flag] = 1; for(int j=1;j

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