LightOJ 1231 Coin Change (I) (背包计数模板题)

1231 - Coin Change (I)

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Time Limit: 1 second(s)

Memory Limit: 32 MB

In a strange shop there are n types of coins of valueA₁, A₂ ... A_n.C₁, C₂,... C_n denote the number of coins of valueA₁, A₂... A_n respectively. You have to find the number of ways you canmakeK using the coins.

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

1112

122

5

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 integersn (1≤ n ≤ 50) andK (1 ≤ K ≤ 1000). The nextline contains2n integers, denotingA₁, A₂ ...A_n, C₁, C₂... C_n (1 ≤ A_i≤ 100, 1 ≤ C_i ≤ 20). AllA_iwill be distinct.

Output

For each case, print the case number and the number of waysKcan be made. Result can be large, so, print the result modulo100000007.

Sample Input	Output for Sample Input
2 3 5 1 2 5 3 2 1 4 20 1 2 3 4 8 4 2 1	Case 1: 3 Case 2: 9

 

题目链接：http://lightoj.com/volume_showproblem.php?problem=1231

题目大意：n种面值为a[i]的硬币，每种对应c[i]个，求能组成金额K的方案数。

解题思路：背包计数 ，dp[i][j]表示前 i 种硬币能组成面值为 j 的方案个数。每个a[i]可用0-c[i]次。递推方程为：当满足条件j-k*a[i]>=0时，dp[i][j]=dp[i-1][j]+dp[i-1][j-k*a[i]]。

代码如下：

#include <cstdio>
#include <cstring>
#define mod 100000007;
int a[52],c[52];
int dp[52][1005];
int main()
{
    int t,cnt=0,n,v,i,j,k;
    scanf("%d",&t);
    while(t--)
    {
		memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&v);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
            scanf("%d",&c[i]);
		dp[0][0]=1;
        for(i=1;i<=n;i++)
            for(j=0;j<=v;j++)
                for(k=0;k<=c[i];k++)
                {
                    if(j-k*a[i]>=0)
                        dp[i][j]=(dp[i][j]+dp[i-1][j-k*a[i]])%mod;
                }
		printf("Case %d: %d\n",++cnt,dp[n][v]);
    }
    return 0;
}