Dead Fraction [POJ1930]

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....，循环的部分不一定是最后一位。

Sample Input

0.2...
0.20...
0.474612399...
0

Sample Output

2/9
1/5
1186531/2500000

 #include<set>
 #include<map>
 #include<queue>
 #include<stack>
 #include<cmath>
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define RG register int
 #define rep(i,a,b)    for(RG i=a;i<=b;++i)
 #define per(i,a,b)    for(RG i=a;i>=b;--i)
 #define ll long long
 #define inf (1ll<<60)
 using namespace std;
 ll n,mx,mn,len;
 char s[];
 inline int read()
 {
     int x=,f=;char c=getchar();
     while(c<''||c>''){if(c=='-')f=-;c=getchar();}
     while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
     return x*f;
 }

 ll gcd(ll a,ll b){    return (a%b==)?b:gcd(b,a%b);}

 void work()
 {
     rep(i,,len+)
     {
         ll A=n,B=,C=i-;ll K=len-C;
         rep(j,,i-)    B=B*+s[j]-'';
         ll X=A-B,Y=pow(,C)*(pow(,K)-1ll);
         ll g=gcd(X,Y);
         if(Y/g<mn)    mx=X/g,mn=Y/g;
     }
     printf("%lld/%lld\n",mx,mn);
 }

 int main()
 {
     while()
     {
         scanf("%s",s);n=len=,mn=inf;
         if(strlen(s)==)    break;
         for(RG i=;;i++)
         {
             if(s[i]=='.') break;
             n=n*+s[i]-'',len++;
         }
         work();
     }
     return ;
 }