POJ 1930 Dead Fraction

 

Dead Fraction

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 1198		Accepted: 354

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

 

 

/* 这题主要是给出一个无限循环小数，要求去求他所对应的具有最小分母的分数。无限循环小数转换为分数有 一定的规律和方法可循： (1)假设输入小数为0.i1 i2 ... it j1 j2 ... jk,即j1 j2 ... jk是循环的部分 那么这个分数可以分两部分来计算即不循环的部分和循环的部分. (2)先来看不循环的部分,这个非常简单.即为 i1i2 ... it / 10 ^ t,即不循环部分的分子分母分别为: upd = i1 i2 ... it; downd = 10 ^ t; (3)再来看循环部分,这部分有点难.在数论中的典型做法是: 取分子up为:j1 j2 ... jk 分母down为: 10 ^ (t + k) - 10 ^ t (4)那么最后所得的分数即为: upd / downd + up / down = (upd * down + downd * up) / (down * up) (5)注意计算的时候要约分.另外由于题目没有明确给出循环部分的长度,所以需要自己枚举,然后计算出 所有的分数取分母最小的.另外要注意的是输入中含有全0的情况(输出0/1),这个太恶心了,因为题目明确说 了输入不全为0的,害我WA了好几次. */ #include <iostream> #include <cmath> #include <cstring> using namespace std; char input[20]; int spos, epos; __int64 minup, mindown; __int64 gcd(__int64 a, __int64 b) { if(b == 0) return a; else return gcd(b, a % b); } //利用最大公约数来约分 void trim(__int64 &up, __int64 &down) { __int64 gcdval; while((gcdval = gcd(up, down)) != 1) { up /= gcdval; down /= gcdval; } } //判断是否全0 bool allZero() { for(int i = strlen(input) - 4; i >= 2; i--) if(input[i] != '0') return false; return true; } int main() { while(scanf("%s", input) && strcmp(input, "0") != 0) { //当不全0时 if(!allZero()) { mindown = -1; epos = strlen(input) - 4; //输入字符串数字位的结束位置 spos = 2; //小数点后面的起始位置 __int64 up = 0, down = 0, upd = 0, downd = 0, dupdown; //枚举循环部分的起始位置 for(int t = spos; t <= epos; t++) { upd = up = 0; int k = spos; while(k < t && input[k] == '0') k++; //找到非循环部分第一个不为0的位置 for(; k <= epos; k++) { if(k < t) upd = upd * 10 + int(input[k] - '0'); //统计非循环部分的分子值 if(k >= t) up = up * 10 + int(input[k] - '0'); //统计循环部分的分子值 } downd = pow(10.0, t - spos); //非循环部分的分母值 down = pow(10.0, epos - spos + 1); //所有小数部分的分母值 dupdown = pow(10.0, epos - t + 1); //循环部分的分母值 down = down - down / dupdown; //制造循环部分小数的最终分母值 if(up != 0) trim(up, down); //约分 if(upd != 0) trim(upd, downd); //约分 __int64 newup, newdown; newup = up * downd + down * upd; newdown = down * downd; if(newup != 0) trim(newup, newdown); //计算最终结果 if(mindown == -1 || newdown < mindown) //取分母最小的 { mindown = newdown; minup = newup; } } printf("%I64d/%I64d/n", minup, mindown); } //全为0的情况 else printf("0/1/n"); } return 0; }