Ural 1011. Conductors

最新推荐文章于 2017-10-03 21:12:14 发布

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最新推荐文章于 2017-10-03 21:12:14 发布

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原文链接：http://www.cnblogs.com/BlackStorm/p/4267597.html

本文探讨了一个关于Ekaterinburg城市中导乘员百分比的问题，并提供了两种算法实现方式来确定该城市的最小市民数量。两种算法分别采用了数学函数和整数转换的方法，通过不断迭代寻找符合条件的最小数值。

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1011. Conductors

Time limit: 2.0 second
Memory limit: 64 MB

Background

Everyone making translations from English to Russian knows an English phrase "Naked conductor runs along the bus". It has two very different meanings.

Problem

Every bus in the Ekaterinburg city has a special man (or woman) called conductor. When you ride the bus, you have to give money to the conductor. We know that there are more than P% conductors and less than Q% conductors of all citizens of Ekaterinburg. Your task is to determine a minimal possible number of Ekaterinburg citizens. By percentage, we know that there are more than P% conductors and less than Q% conductors of all Russian citizens in this city

Input

Two numbers P, Q such that 0.01 ≤ P, Q ≤ 99.99. Numbers are given with 2 digits precision. These numbers are separated by some spaces or "end of line" symbols.

Output

The minimal number of Ekaterinburg citizens.

Sample

input	output
13 14.1	15

Notes

If there are 15 citizens and 2 conductors among them in Ekaterinburg, then there are 13 1/3 % conductors of all citizens.

Problem Source: USU Championship 1997

来源： <http://acm.timus.ru/problem.aspx?space=1&num=1011>

初步想法：从 1 开始遍历，当 n/p 和 n/q 之间存在整数，则所夹最小整数就是所求结果了。但下面的程序 floor 、 ceil 、 fabs 太耗时间了，虽然在自己电脑上都是秒出，但在评测机上达到了 2s 多，华丽丽 TLE。

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

int main(){
    long n=0;
    double p,q,tp,tq;
    scanf("%lf%lf",&p,&q);
    p/=100.0;q/=100.0;
    while(++n){
        tp=floor(n/p);
        tq=ceil(n/q);
        if(fabs(tq-tp)<1E-2) break;
        //if(tp>tq) break;
        //if(tp<tq) continue;
    }
    printf("%.0f",tp);
    return 0;
}

改进：整个算法就循环最耗时间了，有针对性地修改即可，直接 long 强转只保留整数部分，于是考虑：

n/p n/q 强转 -> tp tq

2.x 1.y 2 1

3.x 1.y 3 1

2.x 2.y 2 2

可以看到上面就是小数直接夹着整数的情况成立条件是 tp > tq 。

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

const double e=1E-6;
int main(){
    long n=0;
    double p,q;
    long tp,tq;
    scanf("%lf%lf",&p,&q);
    p/=100.0;q/=100.0;
    while(++n){
        //tp=long(n/p);  //WA
        //tq=long(n/q);
        tp=long(n/p-e);
        tq=long(n/q+e);
        if(tq<tp) break;
    }
    printf("%ld\n",tq+1);
    return 0;
}