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  • sicily 6774. Buying Mortadella

    模拟赛最后一题,看懂了也是挺水的,难度算倒数第二了吧……
    Description
    The National Supermarket Chain (NSC) likes to boast that it has the lowest price for mortadella in the country. In fact, if a customer manages to find cheaper mortadella in any other chain, the NSC will match the price for that customer.
    Matej and Filip decided to accept that challenge. They will visit N different supermarket chains in order to find mortadella not only cheaper than the one in NSC, but the cheapest on the market. If they are successful, they will be able to buy the cheapest mortadella in an NSC branch close to their school.
    NSC was hoping that no one would be able to find cheaper mortadella since all supermarket chains (including NSC) express mortadella prices in a convoluted way: X dollars for Y grams of mortadella.
    Write a program to, given mortadella prices in NSC as well as the remaining N chains, determine the price that Matej and Filip will have to pay for 1000 grams of mortadella in the NSC close to their school.

    Input
    The first line of input contains two positive integers XNSC (1 ≤ XNSC ≤ 100) and YNSC (1 ≤ YNSC ≤ 1000), where XNSC is the price of YNSC grams of mortadella in the NSC chain.
    The second line of input contains the positive integer N (1 ≤ N ≤ 100), the number of supermarket chains (excluding NSC).
    Each of the following N lines contains two positive integers Xi (1 ≤ Xi ≤ 100) and Yi (1 ≤ Yi ≤ 1000), i=1..N, where Xi is the price of Yi grams of mortadella in the ith supermarket chain.

    Output
    The first and only line of output must contain the requested real number (price), rounded to two decimal places.

    View Code
     1 #include<stdio.h>
     2 int main()
     3 {
     4     double x, y;
     5     double min;
     6     int n;
     7     
     8     scanf("%lf %lf", &x, &y);
     9     min = x / y;
    10     
    11     scanf("%d", &n);
    12     
    13     while(n--)
    14     {
    15         scanf("%lf %lf", &x, &y);
    16         if( x/y < min )
    17         {
    18             min = x/y;
    19         }
    20     }
    21     
    22     printf("%.2lf\n", min*1000);
    23     
    24     return 0;
    25 }
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  • 原文地址:https://www.cnblogs.com/joyeecheung/p/2813732.html
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