ACMer
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1095 Accepted Submission(s): 466
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.
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
题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1701
解题：
好久没做题，好久没写解题思路了，连简答题都不会做了，下午的月赛挂了，我悲剧了，在此发牢骚了。o(︶︿︶)o唉
好好的一题简答题，硬是WA，浮点型的陷阱，不止一次掉进去了。刚开始看不懂题目，ACMer人数在P%和Q%之间，那么P%的人与Q%的人就不能相同，因为一个是至少，一个是至多。看代码：
#include <iostream>
using namespace std;
int main()
{
int t,i;double a,b;
cin>>t;
while (t--)
{
cin>>a>>b;
for(i=1;;i++)
{
if ((int)(a*i/100)<(int)(b*i/100))
break;
}
cout<<i<<endl;
}
return 0;
}
运行浮点型要特别注意，下面的情况经常出现。
