POJ-1742-多重背包-Coins

最新推荐文章于 2020-10-10 19:38:54 发布

lele_pipi

最新推荐文章于 2020-10-10 19:38:54 发布

阅读量403

点赞数

CC 4.0 BY-SA版权

分类专栏：背包问题文章标签：背包问题多重背包二进制

本文链接：https://blog.youkuaiyun.com/lele_pipi/article/details/38318199

背包问题专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一个经典的完全背包问题，通过示例详细解释了如何使用多种背包算法（包括完全背包、0-1背包和多重背包）来解决实际问题，并提供了一段C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Coins

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box 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

题意:有面值为A1,A2,,,An的钱,数量分别为C1,C2,,,Cn,问刚好能组成1~m中的数有几个,详解点这里!

代码大致相同 .

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int dp[200000];
int m;
void cpl(int c,int w)//完全背包
{
    int i;
    for(i=c;i<=m;i++)
    {
        if(dp[i]<dp[i-c]+w)
            dp[i]=dp[i-c]+w;
    }
}
void one(int c,int w)//0-1背包
{
    int i;
    for(i=m;i>=c;i--)
    {
        if(dp[i]<dp[i-c]+w)
            dp[i]=dp[i-c]+w;
    }
}
void muty(int c,int w,int amount)//多重背包
{
    int v;
    if(c*amount>=m)             //相当于物体对于容量为sum/2的背包,个数无限多,所以当成完全背包处理
    {
        cpl(c,w);
        return;
    }
    v=1;
    while(v<=amount)        //转化成0-1背包处理,用二进制压缩进行优化
    {
        one(c*v,w*v);
        amount-=v;
        v*=2;
    }
    one(c*amount,amount*w);
}
int main()
{
    int n,A[105],C[105],i,j,cnt;
    while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
    {
        cnt=0;
        for(i=1;i<=n;i++)
            scanf("%d",&A[i]);
        for(i=1;i<=n;i++)
            scanf("%d",&C[i]);
            memset(dp,0,sizeof(dp));
            for(j=1;j<=n;j++)
            {
                muty(A[j],A[j],C[j]);
            }
            for(i=1;i<=m;i++)
            {
                if(dp[i]==i)
                cnt++;
            }
        printf("%d\n",cnt);
    }

}