1231 - Coin Change (I)

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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.

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;
}