Coins （混合背包）

最新推荐文章于 2024-05-21 10:00:00 发布

Mannix_Y

最新推荐文章于 2024-05-21 10:00:00 发布

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CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/Mannix_Y/article/details/80997449

本文介绍了一种通过编程解决特定硬币组合支付问题的方法。该问题要求计算在给定不同面额硬币数量的情况下，可以构成多少种不同的支付金额。通过使用动态规划技术，文章提供了一个C++程序实现，用于找出所有可能的支付组合。

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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

8
4

题意：价值为Ai的物品有Ci个，问选取其中任意物品组成小于m的价值，问这样的价值最多有多少个

//1563ms暴力飘过，混合背包的方法网上有很多就不写了
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define MT(a,b) memset(a,b,sizeof(a))
#define ll long long
const int maxn = 2e5+5;
const int ONF = 0x3f3f3f3f;
const int INF = -0x3f3f3f3f;
const int N = 1e3+5;

int main()
{
    int l=0;
    while(!0)
    {
        int n,p;cin>>n>>p;
        if(!(n+p))break;
        int a[N],num[N];
        MT(a,0);MT(num,0);
        int cnt=0;
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        for(int i=0;i<n;i++)scanf("%d",&num[i]);
        int f[maxn];MT(f,0);
        f[0]=1;
        int ans = 0;
        for(int i=0;i<n;i++)
        {
            int che[maxn];MT(che,0);
            for(int j=a[i];j<=p;j++)
            {
                if(f[j-a[i]]&&!f[j])
                {
                    if(che[j-a[i]]<num[i])
                    {
                        ans++;
                        f[j]=f[j]|f[j-a[i]];
                        che[j]=che[j-a[i]]+1;
                    }
                }
            }
        }
        cout<<ans<<endl;
    }
	return 0;
}