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[20];
inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

ll gcd(ll a,ll b){    return (a%b==0)?b:gcd(b,a%b);}

void work()
{
    rep(i,2,len+1)
    {
        ll A=n,B=0,C=i-2;ll K=len-C;
        rep(j,2,i-1)    B=B*10+s[j]-'0';
        ll X=A-B,Y=pow(10,C)*(pow(10,K)-1ll);
        ll g=gcd(X,Y);
        if(Y/g<mn)    mx=X/g,mn=Y/g;
    }
    printf("%lld/%lld
",mx,mn);
}

int main()
{
    while(1)
    {
        scanf("%s",s);n=len=0,mn=inf;
        if(strlen(s)==1)    break;
        for(RG i=2;;i++)
        {
            if(s[i]=='.') break;
            n=n*10+s[i]-'0',len++;
        }
        work();
    }
    return 0;
}