zoukankan      html  css  js  c++  java
  • Codeforces Round #583 (Div. 1 + Div. 2, based on Olympiad of Metropolises) A题

    A. Optimal Currency Exchange

    Problem Description:

    Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
    Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.

    Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.

    Input
    The first line of the input contains one integer n (1≤n≤108) — the initial sum in rubles Andrew has.

    The second line of the input contains one integer d (30≤d≤100) — the price of one dollar in rubles.

    The third line of the input contains integer e (30≤e≤100) — the price of one euro in rubles.

    Output
    Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.

    Input

    100
    60
    70

    Output

    40

    Input

    410
    55
    70

    Output

    5

    题意:有n个卢布,要换成美元和欧元,使手上剩余的卢克最少。一美元价值d卢布,一欧元价值e卢克.

    思路:关键:欧元的面值最小为5,则是5的倍数,美元最小面值为1。

    写法:特判+枚举。

    AC代码:

    //欧元的面值最小为5,其实的都少5的倍数,美元最小面值为1。
    /*
    1 dollar---> d rubles
    1 euro ----> e rubles
     
    dollar  1 2 5 10 20 50 100
    euro     5 10 20 50 100 200
     
    */
    #include<bits/stdc++.h>
     
    using namespace std;
     
    #define int long long
     
    signed main(){
        int n,d,e;
        cin>>n>>d>>e;
        int ans=n;
        int x1=n/e;
        if(x1<5){
            printf("%lld
    ",n-(n/d)*d);return 0;
        }else{
            int t1=n/(e*5);
            for(int i=0;i<=t1;i++){
                int temp=n;
                temp=temp-e*i*5-((n-e*i*5)/d)*d;
                //int sum2=temp-(temp/d)*d;
                ans=min(ans,temp);
            }
            printf("%lld
    ",ans);
            
        }
        return 0;
    }
  • 相关阅读:
    table布局与div布局
    HTML一般标签
    jquery
    PDO对象
    分页例题
    投票练习
    封装 链接数据库类
    访问数据方法
    面相对象多态
    面向对象
  • 原文地址:https://www.cnblogs.com/pengge666/p/11474409.html
Copyright © 2011-2022 走看看