lightoj1231--Coin Change (I)(简单dp，背包计数)

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 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 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

2

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

题意：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

思路很简单，就是枚举第i个物体放入Ci个的情况下，用前i-1个物体去填补M-Ci*Ai体积的值累和

但写法有两种，递归与非递归。思路上没有区别，但非递归在复杂度系数上较小

先给出两种代码：

递归：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
using namespace std;
typedef long long ll;
int d[55][1005];
int a[55];
int b[55];
int dp(int n,int m)
{
    if(d[n][m]>=0)  return d[n][m];
    if(n==0)
    {
        if(m%a[0]==0&&m/a[0]<=b[0])return  d[n][m]=1;
        else return d[n][m]=0;
    }
    int res=0;
    for(int i=0;i<=b[n]&&i*a[n]<=m;i++)
        res+=dp(n-1,m-i*a[n]);
    return d[n][m]=res%100000007;
}
int main()
{
    int t,fir=1;
    cin>>t;
    while(t--)
    {
        int n,m;
        scanf("%d %d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&b[i]);
        printf("Case %d: ",fir++);
        memset(d,-1,sizeof(d));
        cout<<dp(n-1,m)<<endl;
    }
    return 0;
}

非递归：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
using namespace std;
typedef long long ll;
int a[55];
int b[55];
int dp[55][1005];
int main()
{
    int t,fir=1;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        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=0; j<=b[i]; j++)
                for(int h=k; h>=j*a[i]; h--)
                {
                    dp[i][h]+=dp[i-1][h-j*a[i]];
                    dp[i][h]%=100000007;
                }
                  printf("Case %d: ",fir++);
        cout<<dp[n][k]<<endl;
    }
    return 0;
}

运行时间比较：

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