2616: Cow Frisbee Team(01背包)

2616: Cow Frisbee Team 

时间限制(普通/Java):1000MS/10000MS     内存限制:65536KByte
总提交: 43            测试通过:17

描述

 

After Farmer Don took up Frisbee, Farmer John wanted to join in the fun. He wants to form a Frisbee team from his N cows (1 ≤ N ≤ 2,000) conveniently numbered 1..N. The cows have been practicing flipping the discs around, and each cow i has a rating R_i (1 ≤ R_i ≤ 100,000) denoting her skill playing Frisbee. FJ can form a team by choosing one or more of his cows.

However, because FJ needs to be very selective when forming Frisbee teams, he has added an additional constraint. Since his favorite number is F(1 ≤ F ≤ 1,000), he will only accept a team if the sum of the ratings of each cow in the team is exactly divisible by F.

Help FJ find out how many different teams he can choose. Since this number can be very large, output the answer modulo 100,000,000.

Note: about 50% of the test data will have N ≤ 19.

输入

 

* Line 1: Two space-separated integers: N and F 
* Lines 2..N+1: Line i+1 contains a single integer: R_i

输出

 

* Line 1: A single integer representing the number of teams FJ can choose, modulo 100,000,000.

样例输入

 

4 5
1
2
8
2

样例输出

 3

题目来源

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;

int n,m;
const ll mod=1e8;
const int maxn=2e3+5;
ll dp[maxn][maxn];    //状态表示 前i件物品于是为j的方案书
int arr[maxn];

int main(){
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>arr[i];
        arr[i]%=m;
        dp[i][arr[i]%m]=1;   //初始化
    }
    for(int i=1;i<=n;i++){
        for(int j=0;j<m;j++){
            dp[i][j]=(dp[i][j]%mod+dp[i-1][j]%mod+dp[i-1][(j-arr[i]+m)%m]%mod)%mod;
        }
    }
    cout << dp[n][0] << endl;
    return 0;

}