Codeforces Round #319 (Div. 2)B. Modulo Sum DP
2 seconds
256 megabytes
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).
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.
3 5
1 2 3
YES
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:给你n,m,n个数,让你从中找出任意数的和mod M==0
题解:背包dp
//
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define memfy(a) memset(a,-1,sizeof(a))
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define maxn 1000005
inline ll read()
{
ll x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
//****************************************
int hashs[maxn],a[maxn*];
bool dp[][maxn];
int main()
{ mem(hashs);
int flag=;
int n=read(),m=read();
FOR(i,,n)
{
scanf("%d",&a[i]);
a[i]=a[i]%m;
if(a[i]==)a[i]=m;
}dp[][]=true;
dp[][]=true;
for(int i=;i<=n;i++)
{
for(int j=m;j>=;j--)
{
if(dp[][j])
{
if(j+a[i]==m){
puts("YES");
return ;
}
dp[][(j+a[i])%m]=true;
}
}
memcpy(dp[],dp[],sizeof(dp[]));
}
cout<<"NO"<<endl;
return ;
}
代码君
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