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... 0Sample Output
2/9 1/5 1186531/2500000 题意:很有意思的一道题,,将一个无限循环小数转化成分母最小的精确分数值.... (注意:循环的部分不一定是最后一位,有可能从小数点后面全是循环部分...我因为这个问题WA了2次,题意模糊) 思路:先推导一个数学等式...过程如下(windows画图写的,很不美观>_<): 推导出上述等式,就可以在知道循环部分和非循环后计算出其精确分数值...所以只需要枚举这个数的末端,将其作为循环部分,其余为非循环 部分即可计算出分母最小的精确分数值.... 代码如下: #include<iostream> #include<math.h> using namespace std; int gcd(int a,int b) { if(!a) return b; return gcd(b%a,a); } int main() { char str[100]; int num,k,all,a,b,i,j,mina,minb,l; while(cin>>str&&strcmp(str,"0")) { mina=minb=1000000000; for(i=2,all=0,l=0;str[i]!='.';i++) { all=all*10+str[i]-48; l++; } for(num=all,k=1,i=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) { mina=a/j; minb=b/j; } } cout<<mina<<'/'<<minb<<endl; } return 0; }