描述
As we all know, every decimal can be converted to a fractional number. So you please write a program to do this, not only ordinary decimal, but also circulating decimal, and remember that the result must be fraction in lowest term.
输入
The input consists of multiple test cases. Each test case consists a Pure decimal (decimal without integer part), and the cyclic part should be enclosed buy brackets.And the length of each data would be less than 15 characters.
输出
For each test case, print each fraction in lowest term in a line.
样例输入
0.25 0.(5) 0.32(692307)样例输出
1/4 5/9 17/52题目来源
ben
题意：小数变分数，括号内的无限循环。 AC代码： #include<iostream> #include<math.h> #include<string> using namespace std; void gcd(__int64 m,__int64 n){ __int64 a=m,b=n,c; while(b!=0){ c=b; b=a%b; a=c; } if(m==0){ printf("0\n"); return; } if(a!=0) printf("%I64d/%I64d\n",m/a,n/a); return; } __int64 powhaha(__int64 l){ __int64 ans=1,a=10; while(l>0){ if(l%2==1) ans*=a; l/=2; a*=a; } return ans; } int main(){ __int64 d,num,wei,len,i,g; string s; while(cin>>s){ num=wei=0; d=g=0; len=s.length(); for(i=2;i<len;i++) if(s[i]=='('){ g++; break; } if(g==0){ for(i=2;i<len;i++) num=num*10+s[i]-48; wei=powhaha(len-2); gcd(num,wei); } else if(s[2]=='('){ for(i=3;i<len-1;i++) num=num*10+s[i]-48; wei=powhaha(len-4)-1; gcd(num,wei); } else{ for(i=2;i<len;i++){ if(s[i]!='('){ wei=wei*10+s[i]-48; d=d*10+9; } else break; } for(i=2;i<len;i++){ if(s[i]==')') break; if(s[i]!='(') num=num*10+s[i]-48; else continue; } num=num-wei; wei=powhaha(len-4)-1; wei=wei-d; gcd(num,wei); } } return 0; }