2844 Coins hdu 一些数字生成小于等于m的数的个数转换成多重背包问题

最新推荐文章于 2021-03-26 11:27:03 发布

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本文介绍了一种解决硬币支付问题的算法实现，通过输入不同面额及数量的硬币，计算在限定金额内可组成的支付金额数目。采用完全背包问题的思路进行优化处理。

Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 695 Accepted Submission(s): 277

Problem Description

Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.

You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[100006],a[200],w[200],num[200];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        if(n==0&&m==0) break;
        for(int i=0;i<n;i++) {cin>>a[i];w[i]=a[i];}
        for(int i=0;i<n;i++) cin>>num[i];
        memset(f,0,sizeof(f));
        for(int i=0;i<n;i++)
        {
            if(a[i]*num[i]>m) //完全背包 
            {
                for(int j=a[i];j<=m;j++) f[j]=max(f[j],f[j-a[i]]+w[i]);
            }
            else
            {
                int l=num[i],k=1;
                while(k<l)
                {
                    for(int j=m;j>=k*a[i];j--) f[j]=max(f[j],f[j-k*a[i]]+k*w[i]);
                    l-=k;
                    k*=2;
                }
                for(int j=m;j>=l*a[i];j--) f[j]=max(f[j],f[j-l*a[i]]+l*w[i]);  
            }            
        }
        int cnt=0;
        for(int i=1;i<=m;i++) 
        {
            //cout<<f[i]<<"..";
            if(f[i]==i) cnt++;
        }//cout<<endl;
        printf("%d/n",cnt);
    }
    return 0;
}