ACMer（数学，有意思）

ACMer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2634    Accepted Submission(s): 1190

Problem Description

There are at least P% and at most Q% students of HDU are ACMers, now I want to know how many students HDU have at least?

 

 

Input

The input contains multiple test cases.
The first line has one integer,represent the number of test cases.
The following N lines each line contains two numbers P and Q(P < Q),which accurate up to 2 decimal places.

 

 

Output

For each test case, output the minumal number of students in HDU.

 

 

Sample Input

1 13.00 14.10

 

 

Sample Output

15

 

 

Source

2007省赛集训队练习赛（1）

 

 

Recommend

lcy

 

 

 

Statistic | Submit | Discuss | Note

设a， b分别是学生总数和acmer人数，则有

a*p/100 <= b <= a*q /100 

由于b是整数，所以a*p /100和a*q/100的整数部分不相同，这样两者间就会至少存在一个整数。

所以我们要求的学生总数ans就是满足floor(ans * q / 100) != floor(ans *p / 100)的最小正整数，ans由1开始取数。

AC CODE：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

int main()
{
    double p, q;
    int ans, T;
    scanf("%d", &T);
    while(T-- && scanf("%lf %lf", &p, &q))
    {
        for(ans = 1; floor(ans * q / 100) == floor(ans *p / 100); ans++){}
        printf("%d\n", ans);
    }
    return 0;
}