codeforces #319 B - Modulo Sum (抽屉原理，dp)

B - Modulo Sum
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
Submit Status

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

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

Hint

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.

 

背包还是理解的不够透彻．．

因为每次都是用那个一维形式的．

这道题的做法类似01背包.

此外还可以有一个优化...

当n>m的时候．．．根绝抽屉原理．．一定为yes..

复杂度可以从o(nm)优到　o(m^2)

 1 /*************************************************************************
 2     > File Name: code/#319/BB.cpp
 3     > Author: 111qqz
 4     > Email: rkz2013@126.com 
 5     > Created Time: 2015年09月15日 星期二 21时31分12秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<iomanip>
10 #include<cstdio>
11 #include<algorithm>
12 #include<cmath>
13 #include<cstring>
14 #include<string>
15 #include<map>
16 #include<set>
17 #include<queue>
18 #include<vector>
19 #include<stack>
20 #include<cctype>
21 #define y1 hust111qqz
22 #define yn hez111qqz
23 #define j1 cute111qqz
24 #define ms(a,x) memset(a,x,sizeof(a))
25 #define lr dying111qqz
26 using namespace std;
27 #define For(i, n) for (int i=0;i<int(n);++i)  
28 typedef long long LL;
29 typedef double DB;
30 const int inf = 0x3f3f3f3f;
31 const int N=1E3+7;
32 int n,m;
33 int a[N];
34 int dp[N][5];
35 int main()
36 {
37   #ifndef  ONLINE_JUDGE 
38 
39   #endif
40     cin>>n>>m;
41     if (n>m)
42     {
43     puts("YES");
44     return 0;
45     }
46     for (int i = 1 ; i <= n ; i++)
47     {
48     int x;
49     scanf("%d",&x);
50     a[i] = x % m;
51     }
52     int x = 0 ;
53     for ( int i = 1 ; i <= n ; i++)
54     {
55     
56     dp[a[i]][1-x] = 1;
57     for ( int j = 1 ; j < m ; j++)
58     {
59         if(dp[j][x])
60         {
61         
62         dp[(j+a[i])%m][1-x] = 1;
63         dp[j][1-x] = 1;
64         }
65     }
66     x = 1-x;
67     if (dp[0][x])
68     {
69         puts("YES");
70         return 0;
71     }
72     
73     }
74     puts("NO");
75     return 0;
76 
77 
78 
79   
80   
81  #ifndef ONLINE_JUDGE  
82   fclose(stdin);
83   #endif
84     return 0;
85 }
View Code

 

posted @ 2015-09-16 11:24  111qqz  阅读(288)  评论(0)  编辑  收藏  举报