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：

复制去Google翻译翻译结果