poj1930(小数化分数）

Dead Fraction

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions:3539		Accepted: 1180

Description

Mike is frantically scrambling to finish his thesis at the last minute. He needs to assemble all his research notes into vaguely coherent form in the next 3 days. Unfortunately, he notices that he had been extremely sloppy in his calculations. Whenever he needed to perform arithmetic, he just plugged it into a calculator and scribbled down as much of the answer as he felt was relevant. Whenever a repeating fraction was displayed, Mike simply reccorded the first few digits followed by "...". For instance, instead of "1/3" he might have written down "0.3333...". Unfortunately, his results require exact fractions! He doesn't have time to redo every calculation, so he needs you to write a program (and FAST!) to automatically deduce the original fractions. 
To make this tenable, he assumes that the original fraction is always the simplest one that produces the given sequence of digits; by simplest, he means the the one with smallest denominator. Also, he assumes that he did not neglect to write down important digits; no digit from the repeating portion of the decimal expansion was left unrecorded (even if this repeating portion was all zeroes).

Input

There are several test cases. For each test case there is one line of input of the form "0.dddd..." where dddd is a string of 1 to 9 digits, not all zero. A line containing 0 follows the last case.

Output

For each case, output the original fraction.

Sample Input

0.2...
0.20...
0.474612399...
0
题目大意是给你一个无限循环的小数，要你找到分母最小的分数，有一个坑点是你不知道从哪里开始这个小数部分开始循环，所以要暴力枚举一下。
我们假设不循环的部分位数为k，循环部分位数为c，那么，让这个小数*pow(10,k+c),然后再减去pow(10,k),就可以把后面无限循环的部分减掉。
设a=all-num,all=pow(10,k+c),num=pow(10,k),那么这个分数就可以表示为a/(pow(10,k+c)-pow(10,k)),求出上下两个的最大公约数进行约分就是最终结果。
代码如下：

#include<iostream>
#include<math.h>
using namespace std;
//#define N 1000
int gcd(int a,int b)
{
    if(b==0)
    return a;
    return gcd(b,a%b);
}
int main()
{
        char str[100];
        int all,num,i,l,k,minb,mina,a,b,j;
        while(cin>>str&&strcmp(str,"0"))
       {
       //int all=0,num=0,i,l=0,k=1,minb=10000000,mina=10000000;
       mina=minb=1000000000;
       for(i=2,all=0,l=0;str[i]!='.';i++)
       {
           //l++;
            all=all*10+str[i]-48;
            l++;
       }
       for(i=1,num=all,k=1;i<=l;i++)
       {
           num/=10;
           k*=10;
           a=all-num;
           b=(int)pow(10.0,l-i)*(k-1);
           j=gcd(a,b);
           if(b/j<minb)
           {
               minb=b/j;
               mina=a/j;
           }
       }
       cout<<mina<<'/'<<minb<<endl;
      }
     return 0;
}