C - Coin Change (III)（多重背包二进制优化）

最新推荐文章于 2023-03-15 13:29:40 发布

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最新推荐文章于 2023-03-15 13:29:40 发布

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原文链接：http://www.cnblogs.com/ERKE/p/3582219.html

本文介绍了一种解决特定硬币组合问题的方法，该问题要求计算使用不同类型的硬币可以产生的不同数值的数量。通过输入不同类型的硬币及其数量，程序能够找出在给定范围内所有可能的值。文章提供了一个具体的代码实现示例。

C - Coin Change (III)

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1233

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_nrespectively. You have to find the number of different values (from 1 to m), which can be produced using these coins.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 100), m (0 ≤ m ≤ 10⁵). The next line contains 2n integers, denoting A₁, A₂... A_n, C₁, C₂ ... C_n (1 ≤ A_i ≤ 10⁵, 1 ≤ C_i ≤ 1000). All A_i will be distinct.

Output

For each case, print the case number and the result.

Sample Input

3 10

1 2 4 2 1 1

2 5

1 4 2 1

Sample Output

Case 1: 8

Case 2: 4

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 #define mod 100000007
 6 int a[1001],c[101],dp[100001],b[101];
 7 int main()
 8 {
 9     int n,k,t,i,j,r;
10     scanf("%d",&t);
11     for(i=1; i<=t; i++)
12     {
13         memset(dp,0,sizeof(dp));
14         scanf("%d%d",&n,&k);
15         for(j=1; j<=n; j++)scanf("%d",&b[j]);
16         for(j=1; j<=n; j++)scanf("%d",&c[j]);
17         int nu=0,re,rs;
18         for(j=1; j<=n; j++)
19         {
20             re=b[j],rs=c[j];
21             c[j]*=b[j];
22             while(rs)
23             {
24                 if(re<c[j])
25                     a[nu++]=re,c[j]-=re;
26                 else a[nu++]=c[j];
27                 rs>>=1;
28                 re<<=1;
29             }
30         }
31         for(j=0; j<nu; j++)
32         {
33             for(r=k; r>=0; r--)
34             {
35                 if(dp[r]&&r+a[j]<=k)
36                     dp[r+a[j]]=1;
37                 if(r==0&&a[j]<=k)
38                     dp[a[j]]=1;
39             }
40 
41         }
42         int sum=0;
43         for(r=0; r<=k; r++)sum+=dp[r];
44         cout<<"Case "<<i<<": ";
45         cout<<sum<<endl;
46     }
47 
48 }