codeforces377B Modulo Sum（抽屉+dp）

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 test(s)
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.

更新----我是傻逼，我是傻逼，我是傻逼，看一个知识点总是粗略的看一下，而后的原理自己都搞的很少。

鸽巢就是告诉了我们n>=m时是一定有的，而我呢？更新看代码吧，，，看过别人的代码

---------完结

也许是自己菜吧，反正觉得这是一道好题。

首先当n>=m时我们直接套用抽屉原理判断，对于n<m的现象自己最先没有想到，看了别人的题解才明白了需要用dp。

dp[i][j]记录的是i个数里，和为j的数字有没有出现过，最后判断dp[n][c]是否等于1

这是自己直接套用的抽屉原理模板，应该可以改简单的，明天再看一下。

/* ***********************************************Author        :MosuCreated Time  :2015/10/10 18:30:45File Name     :\Users\Mosu\Desktop\project\BC.cpp************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7ffffffint neigb[1000009];int dp[1005][1005];long long S;struct Remnant{int h,r;}R[1000009];bool cmp(const Remnant &a,const Remnant &b){if(a.r==b.r)return a.h<b.h;return a.r<b.r;}int main(){int c,n,k,h;while(scanf("%d%d",&n,&c)!=EOF){for(int i=0;i<n;i++)scanf("%d",&neigb[i]);if(n>c){k=-1;S=0;for(int i=0;i<n;i++){//scanf("%d",&neigb[i]);S+=neigb[i];R[i].r=S%c;R[i].h=i+1;if(k==-1&&R[i].r==0)k=i;}if(k==-1){sort(R,R+n,cmp);for(int i=0;i<n-1;i++){if(k==-1&&R[i].r==R[i+1].r){k=R[i].h;h=R[i+1].h;}}if(k==-1)printf("NO\n");else{printf("YES\n");}}else{printf("YES\n");}}else{for(int i=0;i<=n;i++)dp[i][0]=1;for(int i=1;i<=n;i++)for(int j=1;j<=c;j++)dp[i][j]=(dp[i-1][(j+neigb[i-1])%c]||dp[i-1][j]);if(dp[n][c])cout<<"YES"<<endl;elsecout<<"NO"<<endl;}}return 0;}

更新：

/* ***********************************************Author        :MosuCreated Time  :2015/10/10 18:30:45File Name     :\Users\Mosu\Desktop\project\BC.cpp************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7ffffffint dp[1009][1009];int s[1000009];int main(){int n,m;while(scanf("%d%d",&n,&m)==2){int k=1;for(int i=0;i<n;i++){scanf("%d",&s[i]);if(s[i]%m==0)k=0;}if(n>m||k==0){printf("YES\n");continue;}sizeof(dp,0,sizeof(dp));for(int i=0;i<=n;i++)dp[i][0]=1;for(int i=1;i<=n;i++)for(int j=0;j<=m;j++)dp[i][j]=(dp[i-1][(j+s[i-1])%m]||dp[i-1][j]);if(dp[n][m])printf("YES\n");elseprintf("NO\n");}return 0;}