CF div2 319 B

B. Modulo Sum
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

题目的意思是问你能不能在n个数里面选若干个相加使得它们的和能整除m，显然当n>=m时肯定可以，那么当n<m时可以利用dp来做

dp[i][j]表示用前i-1个数的sum值得到j(这里j=(sum%m)),显然状态转移方程为 dp[i+1][(j+a[i])%m] = 1，或者 dp[i+1][j];

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <queue>
 5 #include <vector>
 6 #include <stack>
 7 
 8 using namespace std;
 9 
10 const int M = 10005;
11 const int maxn = 1100000;
12 typedef long long ll;
13 
14 vector<int>G[maxn];
15 queue<int>Q;
16 stack<int>st;
17 
18 priority_queue<int>q;
19 
20 
21 int l[maxn],r[maxn];
22 int a[maxn];
23 int dp[1005][1005];
24 int main()
25 {
26 
27     int n;
28     int ans = 0;
29     int m;
30     scanf("%d%d",&n,&m);
31     for(int i=1;i<=n;i++){
32         scanf("%d",&a[i]);
33 
34     }
35 
36     if(n>=m){
37         printf("YES\n");
38     }
39     else {
40 
41         for(int i=1;i<=n;i++){
42                 dp[i][a[i]%m] = 1;
43             for(int j=0;j<m;j++){
44                 if(dp[i-1][j]){
45                     dp[i][(j+a[i])%m] = 1;
46                     dp[i][j] = 1;
47                 }
48             }
49         }
50         if(dp[n][0])printf("YES\n");
51         else printf("NO\n");
52     }
53 
54 
55 
56     return 0;
57 }
View Code

 

posted @ 2015-09-16 22:17  lmlyzxiao  阅读(117)  评论(0)    收藏  举报