POJ 1930 Dead Fraction(gcd—枚举循环节，无限循环小数变最简分数)


[align=center]Dead Fraction[/align]

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2323		Accepted: 748

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

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...).

题意：把无限循环小数变最简分数，但不确定循环节是哪几位，要求输出分母最小的分数。

题解：枚举循环节，然后不断更新最小的分母，同时记录下对应的分子即可。关于循环小数化分数，在之前的题解中给出过详细的转化方法及证明。链接：小数化分数

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll __int64
using namespace std;

ll gcd(ll n,ll m)
{
ll x;
while(m)
{
x=n%m;
n=m;
m=x;
}
return n;
}

int main()
{
int len,i,j,point,k;
ll inter;
char str[20];
while(scanf("%s",str)&&strcmp(str,"0")!=0)
{
len=strlen(str);
for(i=2;i<len;++i)
{
if(str[i]=='.')
{
point=i;//记录数字后面第一个点出现的位置
break;
}
}
ll ans_num=0x3f3f3f3f,ans_cnt;
for(i=2;i<point;++i)
{
j=i-2;//j表示小数点后非循环节的长度
ll cnt=0,num=0;
for(k=i;k<point;++k)
{
cnt=cnt*10+(str[k]-'0');//分子
num=num*10+9;//分母
}
inter=0;
for(k=2;k<i;++k)
inter=inter*10+(str[k]-'0');//循环节前面的数字
cnt+=inter*num;
for(k=0;k<j;++k)
num*=10;
ll res=gcd(num,cnt);
cnt/=res;
num/=res;
if(num<ans_num)//更新最小的分母
{
ans_num=num;
ans_cnt=cnt;
}
}
printf("%I64d/%I64d\n",ans_cnt,ans_num);
}
return 0;
}