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最新推荐文章于 2019-08-03 22:12:19 发布

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本文介绍了一种使用动态规划解决硬币组合问题的方法，通过给定不同类型的硬币及其数量来找出构成特定金额的所有可能方式。

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 contains2n 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

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>

#define ll __int64
#define lll unsigned long long
#define MAX 1000009
#define eps 1e-8
#define INF 0xfffffff
#define mod 100000007
using namespace std;

int w[MAX];
int c[MAX];
int v;
int n;

/*

题意：给你不同种类的硬币和个数，让你求组成面值为k的方法数

解法：最开始觉得是多重背包，然后一顿找模板，最后发现只是个简单DP。还有啊，对多重背包那里实在是搞得不好，不扎实，昨天打的好渣的说。

*/
int dp[55][1009];//前i种物品组成容量为j的取法

int solve_dp()
{
memset(dp,0,sizeof(dp));
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=c[i];j++)
{
if(j*w[i]<=v)//小于预期额度
dp[i][j*w[i]] = 1;//选择前i种硬币下可以凑成的额度
}
}
for(int i = 2;i<=n;i++)
{
for(int j = 1;j<=v;j++)
{
for(int l = 0;l<=c[i];l++)
{
if(j-l*w[i]>0)
dp[i][j] = (dp[i][j] + dp[i-1][j - l*w[i]]) %mod;//当前的方法数+前i-1种硬币组成j - l*w[i]的额度的方法数
}
}
}
return dp[n][v];
}
int main()
{
int d = 1;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&v);
for(int i = 1;i<=n;i++) scanf("%d",&w[i]);
for(int i = 1;i<=n;i++) scanf("%d",&c[i]);
printf("Case %d: %d\n",d++,solve_dp());
}

return 0;
}