UVA 147 Dollars (子集和问题 & DP)

最新推荐文章于 2021-02-24 11:25:14 发布

wchhlbt

最新推荐文章于 2021-02-24 11:25:14 发布

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分类专栏：动态规划文章标签： uva dp

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本文介绍了一种使用动态规划解决新西兰货币组合问题的方法，通过预处理面额转换为单位值，实现了快速计算任意金额的组成方式。

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题目：New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1×20c, 2×10c, 10c+2×5c, and 4×5c. Input Input will consist of a series of real numbers no greater than $300.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00). Output Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 6), followed by the number of ways in which that amount may be made up, right justified in a field of width 17.

解题思路：首先既然所有的钱数都是五分钱的整数倍（题目保证红字），所以我们把5分当作单位1，300元也就变为6000，所以题目就变成了用1,2,4,10,20,40,100,200,400,1000,2000这些面值的钱币得到一个一个任意小于6000金额的方法数。是一个子集和问题（完全背包）。

设dp[i][j] 表示使用前 i 类币值能够得到金额 j 的方法数，分析可得到下列递推式

dp[i][j] = dp[i][j-a[i]] + dp[i-1][j]

注意：本题需要预处理6000以内的结果，并注意输出格式要求

AC代码：

#include <bits/stdc++.h>
#define maxn 6005
using namespace std;
typedef long long LL;
int a[20] = {0,1,2,4,10,20,40,100,200,400,1000,2000};
LL dp[12][maxn];
int main()
{
    //freopen("test.txt","r",stdin);
    for(int i = 0; i<=6000; i++)
    	dp[1][i] = 1;
	for(int i = 2; i<12; i++)
    {
    	for(int j = 0; j<=6000; j++)
    	{
    		dp[i][j] = dp[i-1][j];
    		if(j>=a[i])
    			dp[i][j] += (dp[i][j-a[i]]);
    	}
    }
    double d;
    cout << fixed << showpoint << setprecision(2);
    while(cin>>d && d!=0.00)
    {
    	int m = int(d*20);
    	cout << setw(6) << d << setw(17) << dp[11][m] << endl;
    }
    return 0;
}

当然还可以更简洁一些

#include <bits/stdc++.h>
#define maxn 6005
using namespace std;
typedef long long LL;
int a[20] = {0,1,2,4,10,20,40,100,200,400,1000,2000};
LL dp[maxn];
int main()
{
    for(int i = 0; i<=6000; i++)
    	dp[i] = 1;
	for(int i = 2; i<12; i++)
    	for(int j = a[i]; j<=6000; j++)
    		dp[j] += dp[j-a[i]];
    double d;
    cout << fixed << showpoint << setprecision(2);
    while(cin>>d && d!=0.00)
    {
    	int m = int(d*20);
    	cout << setw(6) << d << setw(17) << dp[m] << endl;
    }
    return 0;
}