ACMer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4763    Accepted Submission(s): 2232

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） 


思路：

也是水题。。。就是遍历，看这个数×a%和×b%所得的两个小数之间有没有整数。如果有那就说明存在一个解。

WA的点就是对浮点数的计算没搞明白。。。。


我到现在都还没搞明白这两个方法算出来的m2和n2为什么不相同。。第一种办法WA了半小时，改成第二种算法就过了。。


代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		double m,n;
		cin>>m>>n;
		for(int i=1;;i++)
		{
			int m2=m*i/100;
			int n2=n*i/100;
			//cout<<m<<" "<<n<<" "<<m1<<" "<<n1<<" "<<m2<<" "<<n2<<endl;
			if(m2<n2)
			{
				cout<<i<<endl;
				break;
			}
		}
	}
	return 0;
}


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