LightOj 1231 Coin Change (II)（完全背包）

Coin Change (II)

Description

In a strange shop there are n types of coins of value A₁, A₂ ... A_n. You have to find the number of ways you can make K using the coins. You can use any coin at most K times.

For example, suppose there are three coins 1, 2, 5. Then if K = 5 the possible ways are:

11111

1112

122

5

So, 5 can be made in 4 ways.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 100) and K (1 ≤ K ≤ 10000). The next line contains n integers, denoting A₁, A₂ ... A_n (1 ≤ A_i ≤ 500). All A_i will be distinct.

Output

For each case, print the case number and the number of ways K can be made. Result can be large, so, print the result modulo 100000007.

Sample Input

2

3 5

1 2 5

4 20

1 2 3 4

Sample Output

Case 1: 4

Case 2: 108

解题思路：

完全背包。

状态转移方程：dp[i][j] = (dp[i][j]+dp[i-1][j-l*a[i]])%MOD。

优化以后就是： dp[j] = (dp[j]+dp[j-a[i]])%MOD 。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MOD = 100000007;const int maxn = 10005;int dp[maxn];int a[105],b[105];int main(){    int T,t = 1;    scanf("%d",&T);    while(T--){        int n,m;        scanf("%d%d",&n,&m);        for(int i = 1; i <= n; i++)            scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        dp[0] = 1;        for(int i = 1; i <= n; i++){            for(int j = a[i]; j <= m; j++){                dp[j] = (dp[j]+dp[j-a[i]])%MOD;            }        }        printf("Case %d: %d\n",t++,dp[m]);    }    return 0;}