CodeForces 592C The Big Race

公倍数之间的情况都是一样的，有循环节。

注意min(a,b)>t的情况和最后一段的处理。C++写可能爆longlong，直接Java搞吧......

import java.io.BufferedInputStream;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    
    public static BigInteger GCD(BigInteger a,BigInteger b)
    {
        if(b.compareTo(BigInteger.ZERO)==0) return a;
        return GCD(b,a.remainder(b));
    }
    
    public static BigInteger MIN(BigInteger a,BigInteger b)
    {
        if(a.compareTo(b)>=0) return b;
        return a;
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner (new BufferedInputStream(System.in));
        
        BigInteger t=sc.nextBigInteger();
        BigInteger a=sc.nextBigInteger();
        BigInteger b=sc.nextBigInteger();
        
        if(MIN(a,b).compareTo(t)>0)
        {
            System.out.print("1"+"/"+"1");
        }
        else{
            BigInteger gcd=GCD(a,b);
            BigInteger lcm=a.multiply(b).divide(gcd);
            BigInteger k=MIN(a,b).subtract(BigInteger.ONE);
            BigInteger ans=BigInteger.ZERO;
            BigInteger tmp=t.divide(lcm);
            ans=k;
            if(tmp.compareTo(BigInteger.ZERO)>0)
            {
                BigInteger u=tmp.subtract(BigInteger.ONE);
                ans=ans.add(u.multiply(k.add(BigInteger.ONE)));
                ans=ans.add(BigInteger.ONE);
                ans=ans.add(MIN(k,t.subtract(lcm.multiply(tmp))));
            }
            BigInteger fz=ans;
            BigInteger fm=t;
            BigInteger e=GCD(fz,fm);
            fz=fz.divide(e);
            fm=fm.divide(e);
            System.out.print(fz.toString()+"/"+fm.toString());
        }
    }
}