POJ 1742 Coins（dp）-优快云博客

本文介绍了一道关于硬币组合的编程题，旨在求解利用给定种类和数量的硬币能够组成的不超过特定总额的不同价格的数量。通过动态规划的方法实现了算法，并提供了完整的代码实现。

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Language:

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 29278		Accepted: 9916

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

8
4

Source

LouTiancheng@POJ

就是输入钱币种类，数量，问可以组合出来的钱币面值种类数量

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100007

int dp[N],sum[N];
int va[200],all,n,num[200];

int main()
{
    int i;
    while(scanf("%d%d",&n,&all),n+all)
    {
        for(i=0;i<n;i++)
            scanf("%d",&va[i]);

        for(i=0;i<n;i++)
            scanf("%d",&num[i]);

        for(i=0;i<=all;i++)
            dp[i]=0;
        dp[0]=1;

        int j,ans=0;

        for(i=0;i<n;i++)
        {
            for(j=0;j<=all;j++)
                sum[j]=0;      //这里的意思是当前这一种钱币在任何一种钱币面值下都没有使用过

            for(j=va[i];j<=all;j++)
                if(!dp[j]&&dp[j-va[i]]&&sum[j-va[i]]<num[i])
            // 没有组合出来, j-va[i]组合出来  并且j-va[i]这种钱币用的va[i]次数小余num[i]
            {
                dp[j]=1;
                sum[j]=sum[j-va[i]]+1;
                ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}