数论——Dead Fraction（POJ - 1930 ）

数论

——Dead Fraction（POJ - 1930 ）
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

Sample Output

2/9
1/5
1186531/2500000

Hint
Note that an exact decimal fraction has two repeating expansions (e.g. 1/5 = 0.2000… = 0.19999…

解析：数论题
一，纯循环小数化分数：循环节的数字除以循环节的位数个9组成的整数。
例如：0.3333……=3/9=1/3；
二，混循环小数：不循环部分和循环节构成的的数减去不循环部分的差，再除以循环节位数个9和添上不循环位数的0。
例如：0.9545454…………=(954-9)/990=945/990=21/22
注意1.pow一定是double类型传进去，后强制转换
2.这里用到枚举循环节的时候，用k去记录循环的位数（k逐一10，后-1）；并且直接用一个tmp去记录不循环的部分（tmp逐一10）；

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<string>
#include<string.h>
#include<iostream>
#define inf 0x3f3f3f3f
using namespace std;
int ans,fz,fm,n,c;
string s;
int gcd(int a,int b){
   if(b==0) return a;
   return gcd(b,a%b);
}
int main()
{
    while(cin>>s&&s!="0")
    {
        c=0,n=0;
        for(int i=2;s[i]!='.';i++){
            n=n*10+s[i]-'0';c++;
        }
        fm=inf;
        int tmp=n,k=1,a,b;
        for(int i=1;i<=c;i++){
            k*=10;
            tmp/=10;
            a=n-tmp;
            b=(int)pow(10.0,c-i)*(k-1);
            int d=gcd(a,b);
            if(fm>b/d){
                fm=b/d;
                fz=a/d;
            }
        }
        printf("%d/%d\n",fz,fm);
    }
    return 0;
}