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 51 2 3
Output
YES
Input
1 65
Output
NO
Input
4 63 1 1 3
Output
YES
Input
6 65 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: [email protected]
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 }