POJ 3085 Quick Change (贪心)

最新推荐文章于 2021-09-06 12:41:26 发布

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本文介绍了一种解决零钱找补问题的算法，通过输入金额，程序计算并输出应找回的硬币数量，包括四分之一美元、十分之一美元、五分之一美元和一分美元。

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Quick Change

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5801		Accepted: 4175

Description

J.P. Flathead’s Grocery Store hires cheap labor to man the checkout stations. The people he hires (usually high school kids) often make mistakes making change for the customers. Flathead, who’s a bit of a tightwad, figures he loses more money from these mistakes than he makes; that is, the employees tend to give more change to the customers than they should get.

Flathead wants you to write a program that calculates the number of quarters ($0.25), dimes ($0.10), nickels ($0.05) and pennies ($0.01) that the customer should get back. Flathead always wants to give the customer’s change in coins if the amount due back is $5.00 or under. He also wants to give the customers back the smallest total number of coins. For example, if the change due back is $1.24, the customer should receive 4 quarters, 2 dimes, 0 nickels, and 4 pennies.

Input

The first line of input contains an integer N which is the number of datasets that follow. Each dataset consists of a single line containing a single integer which is the change due in cents, C, (1 ≤ C ≤ 500).

Output

For each dataset, print out the dataset number, a space, and the string:

Q QUARTER(S), D DIME(S), n NICKEL(S), P PENNY(S)

Where Q is he number of quarters, D is the number of dimes, n is the number of nickels and P is the number of pennies.

Sample Input

Sample Output

1 4 QUARTER(S), 2 DIME(S), 0 NICKEL(S), 4 PENNY(S)
2 1 QUARTER(S), 0 DIME(S), 0 NICKEL(S), 0 PENNY(S)
3 7 QUARTER(S), 1 DIME(S), 1 NICKEL(S), 4 PENNY(S)

Source

Greater New York 2006

就是找零钱，无需多言。

代码：

#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
int main()
{
int n,m,i;
int f[4]={25,10,5,1};
while(scanf("%d",&n)!=EOF&&n)
{
for( i=1;i<=n;i++)
{
int k[4]={0};
scanf("%d",&m);
while(m>=f[0]) {m-=f[0];k[0]++;}
while(m>=f[1]) {m-=f[1];k[1]++;}
while(m>=f[2]) {m-=f[2];k[2]++;}
while(m>=f[3]) {m-=f[3];k[3]++;}
printf("%d %d QUARTER(S), %d DIME(S), %d NICKEL(S), %d PENNY(S)\n",i,k[0],k[1],k[2],k[3]);
}
}
return 0;
}