LightOj 1231 Coin Change (I)（部分背包）

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本文介绍了一个经典的背包问题——CoinChange，并提供了一种有效的动态规划解决方案。问题旨在利用不同面额的硬币组合来构成给定的目标金额，探讨了如何计算可能的组合数。文章还分享了AC代码实现，帮助读者理解算法的具体应用。

Coin Change (I)

Description

In a strange shop there are n types of coins of value A₁, A₂ ... A_n. C₁, C₂, ... C_n denote the number of coins of value A₁, A₂ ... A_n respectively. You have to find the number of ways you can make K using the coins.

For example, suppose there are three coins 1, 2, 5 and we can use coin 1 at most 3 times, coin 2 at most 2 times and coin 5 at most 1 time. Then if K = 5 the possible ways are:

1112

122

So, 5 can be made in 3 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 ≤ 50) and K (1 ≤ K ≤ 1000). The next line contains 2n integers, denoting A₁, A₂ ... A_n, C₁, C₂ ... C_n (1 ≤ A_i ≤ 100, 1 ≤ C_i ≤ 20). 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

3 5

1 2 5 3 2 1

4 20

1 2 3 4 8 4 2 1

Sample Output

Case 1: 3

Case 2: 9

解题思路：

给你n个物品的体积和数量，让你求有多少种组合能恰好装满m体积的背包。

用dp[i][j]表示前i种物品，组成j的容量有几种组法，则状态转移方程为：

dp[i][j] += dp[i-1][j-k*a[i]];

枚举第i种物品的时候可以取一个，取两个，。。，当然最后别忘了：也可以不取。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int MOD = 100000007;
const int maxn = 1005;
int dp[55][maxn];
int a[55],b[55];

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]);
        for(int i = 1; i <= n; i++)
            scanf("%d",&b[i]);
        memset(dp,0,sizeof(dp));
        dp[0][0] = 1;
        for(int i = 1; i <= n; i++){
            for(int j = m; j >= 0; j--){
                for(int k = 1; k <= b[i]; k++) {
                    if(j-k*a[i] >= 0)
                        dp[i][j] += dp[i-1][j-k*a[i]];
                }
            }
            for(int j = 0; j <= m; j++)
                dp[i][j] = (dp[i][j]+dp[i-1][j])%MOD;
        }
        printf("Case %d: %d\n",t++,dp[n][m]);
    }
    return 0;
}