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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
3 5
1 2 3
YES
1 6
5
NO
4 6
3 1 1 3
YES
6 6
5 5 5 5 5 5
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 }
codeforces #319 B - Modulo Sum (抽屉原理，dp)
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原文地址：http://www.cnblogs.com/111qqz/p/4812759.html
