POJ 3085:Quick Change

最新推荐文章于 2021-09-06 12:41:26 发布

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最新推荐文章于 2021-09-06 12:41:26 发布

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原文链接：http://www.cnblogs.com/im0qianqian/p/5989583.html

本文介绍了一个名为QuickChange的程序，该程序旨在帮助解决零售业中常见的找零错误问题。通过对输入的找零金额进行计算，程序能够快速准确地给出所需的硬币组合，包括25美分的硬币（Quarter）、10美分的硬币（Dime）、5美分的硬币（Nickel）和1美分的硬币（Penny）。这不仅减少了员工的工作负担，也确保了找零过程的准确性。

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Quick Change

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6111		Accepted: 4358

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

你离开了，我的世界里只剩下雨。。。

#include <stdio.h>
int main()
{
    int n,c,q,d,x;
    int i;
    scanf("%d", &n);
    for (i = 1; i <= n; i++)
    {
        q = 0, d = 0, x = 0;
        scanf("%d", &c);
        q = c / 25;
        c = c % 25;
        d = c / 10;
        c = c % 10;
        x = c / 5;
        c = c % 5;
        printf("%d %d QUARTER(S), %d DIME(S), %d NICKEL(S), %d PENNY(S)\n", i, q, d, x, c);
    }
    return 0;
}

转载于:https://www.cnblogs.com/im0qianqian/p/5989583.html