hdu-2844 Cions-多重背包

最新推荐文章于 2021-03-09 15:37:20 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，通过多种背包问题的变体（如01背包、完全背包和多重背包），计算在给定硬币种类及数量的情况下，能够支付的不同价格数目。

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Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18336 Accepted Submission(s): 7208

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

Source

2009 Multi-University Training Contest 3 - Host by WHU

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#include<stdio.h>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int f[1000010];//满足价值不超过价值 i的最大价值 
int a[105],c[105];
int n,m;

//m-背包容量，v-物品体积，w-物品价值，num-物品数量
//01背包 
void OneZeroPack(int m,int v,int w){
	for(int i=m;i>=v;i--){
		f[i]=max(f[i],f[i-v]+w);
	} 
} 
//完全背包 
void CompletePack(int m,int v,int w){
	for(int i=v;i<=m;i++){
		f[i]=max(f[i],f[i-v]+w);
	}
}
//多重背包  
void MutiplePack(int m,int v,int w,int num){
	if(v*num>=m){
		CompletePack(m,v,w);
		return ;
	} 
	int k=1;
	for(k=1;k<=num;k<<=1)
	{
		OneZeroPack(m,k*v,k*w);
		num-=k; 
	}
	if(num)
	  OneZeroPack(m,num*v,num*w);
}
int main()
{
	while(~scanf("%d%d",&n,&m)&&(n||m))
	{
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(int i=0;i<n;i++)
			scanf("%d",&c[i]);
		for(int i=0;i<=m;i++)
			f[i]=-inf;
		f[0]=0;
		for(int i=0;i<n;i++)
		{
			MutiplePack(m,a[i],a[i],c[i]);
		}
		int sum=0;
		for(int i=1;i<=m;i++)
			if(f[i]>0)
			sum++;
		printf("%d\n",sum);
	}
	return 0; 
}