POJ 1930 Dead Fraction【数学】

Dead Fraction

Time Limit: 1000MS	Memory Limit: 30000K
Total Submissions: 2463	Accepted: 802

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...).

Source

Waterloo local 2003.09.27

题意：

英语实在不行，题意都没读明白，只知道是把小数变成分数，但是却没发现题目都给出循环节，需要自己枚举去找.........

另外输出的是所有的可能性中最简形式下分母最小的那一项...

题解：

循环小数化简为分数的题解请点此处

唯一的变化，就是外加一层循环，枚举循环节开始的地点，并记录出现的最简形式的分母的最小值

注意题目中的某个坑，题目说好的没这样的情况，结果...

/*
http://blog.csdn.net/liuke19950717
*/
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
	if(!b)
	{
		return a;
	}
	return gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
	return a/gcd(a,b)*b;
}
int zero(char s[],int len)
{
	for(int i=2;i<len-3;++i)
	{
		if(s[i]!='0')
		{
			return 0;
		}
	}
	return 1;
}
int main()
{
	char s[105];
	while(~scanf("%s",s))
	{
		if(s[0]=='0'&&s[1]==0)
		{
			break;
		}
		ll len=strlen(s);
		if(zero(s,len))//玩这样的坑有意思吗？ 
		{
			printf("0/1\n");
			continue;
		}
		ll ansx,ansy=1<<31-1;
		for(ll pos=2;pos<len-3;++pos)
		{
			ll a=0,x=1;//不循环部分 
			for(ll i=2;i<pos;++i)
			{
				a=a*10+s[i]-'0';
				x*=10;
			}
			ll b=0,y=0;//循环部分 
			for(ll i=pos;i<len-3;++i)
			{
				b=b*10+s[i]-'0';
				y=y*10+9;
			}
			y*=x;
			ll ty=lcm(x,y),tx=a*(ty/x)+b*(ty/y);
			ll tp=gcd(tx,ty);
			tx/=tp;ty/=tp;
			if(ty<ansy)
			{
				ansx=tx;ansy=ty;
			}
		}
		printf("%lld/%lld\n",ansx,ansy);
	}
	return 0;
}