POJ 1276 Cash Machine 多重部分和模板题 动态规划

本文探讨了一种ATM机现金分配算法,旨在通过使用不同面额的纸币组合,实现对用户请求现金金额的有效响应。算法采用动态规划方法,考虑了每种面额纸币的可用数量,以确定能提供的最大现金总额,不超过用户请求的金额。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Description
A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example,
N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10
means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.
Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.
Notes:
@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.
Input
The program input is from standard input. Each data set in the input stands for a particular transaction and has the format:
cash N n1 D1 n2 D2 … nN DN
where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct.
Output
For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.
Sample Input
735 3 4 125 6 5 3 350
633 4 500 30 6 100 1 5 0 1
735 0
0 3 10 100 10 50 10 10
Sample Output
735
630
0
0
Hint
The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash.
In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash.
In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.

题目大意:
给出n种不同面额的硬币以及它们的数量,现在的目的是凑出不大于给定cash的最大面额。

解题思路:
多重部分和问题的模板题,具体分析见我的上一篇博客
POJ 1742 Coins 动态规划 多重部分和问题 解题模板

AC代码:

#include<iostream>
#include<stdio.h>
using namespace std;
int a[11]; int c[11]; //a[i]表示面额 c[i]表示数量
int dp[100001];
bool can_cash[100001];
int main() {
	int cash, n;
	while (cin >> cash >> n) {
		for (int i = 1; i <= n; i++) {
			scanf("%d%d", &c[i], &a[i]);
		}
		if (cash == 0 || n == 0) {
			cout << 0 << endl;
			continue;
		}
		for (int i = 1; i <= cash; i++) { can_cash[i] = false; }
		can_cash[0] = true;
		for (int i = 1; i <= n; i++) {
			for (int i = 1; i <= cash; i++) { dp[i] = 0; }
			for (int j = a[i]; j <= cash; j++) {
				if (!can_cash[j] && can_cash[j - a[i]] && dp[j - a[i]] < c[i]) {
					can_cash[j] = true;
					dp[j] = dp[j - a[i]] + 1;
				}
			}
		}
		for (int k = cash; k >= 0; k--) {
			if (can_cash[k]) {
				cout << k << endl;
				break;
			}
		}
	}
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值