Gym

(Gym.cpp/c/pas)

Topic Description

Muji finally arrives at the gym, VAN 様's lair, but he is already alone. He decided to have a showdown with VAN. He decided to play ♂run♂step with VAN . Known track length $ l $m, and Muji can run one step and can only run$ n $m, VAN can run one step and can only run$ m $meter. Now it is stipulated that the runner cannot run out$ k $m. And whoever runs farther in the end wins. To be fair, $ k$ is a completely random positive integer between $1 $ to $ l $. Now Muji wants to know what is the probability of a draw with VAN 様.

Input Description (gym.in) Input Description

The first line is three integers, followed by $l,n,m$ ;

Output Description (gym.out) Output Description

A reduced true score, in the format a/b, which is the probability of a draw between Muji and VAN

Sample Input

10 3 2

Sample Output

3/10

Sample Interpretation

When $k$ is $1,6,7$, Muji will tie with VAN 様

Data Range Data Size

For 30% of the data, $n,m,l\le 10^6$

For 100% data, $n,m,l\le 5\times {10}^{18}$

answer

First determine that this is a number thesis, so I think in this direction.

Obviously, the necessary and sufficient conditions for a draw between Muji and VAN are: $k\mod n=k\mod m$ .

It is not difficult to find that when $n=m$ , $k$ obviously holds. The above formula will report an error, so you need to make a special judgment:

if(n==m)
{
    fout<<"1/1";
    return 0;
}

Then go on to start the happy derivation...

Convert the $k$ in the above formula into the formula with remainder, there are $k-t_1 n=k-t_2 m$ , $t_1 n=t_2 m$ .

Let the total number of $k$ satisfying the tie between Muji and VAN be $ans$ .

It is not difficult to find that when $k=\infty$ , Muji and VAN meet for the first time at $[a,b]$ . So, when $k<[a,b]$ , $ans=\min(m,n)$ (neither of them took a step).

Following this line of thinking, we find that two people are equivalent at $[a,b]$ at the starting point. So it is not difficult for us to come up with a correct solution.

The cancer is that $[a,b]$ unsigned long long can't be stored... So long double must be used if necessary .

code

#include <fstream>
#include <algorithm>

using namespace std;

typedef long long LL;

LL l,n,m;

int main()
{
    ifstream fin("gym.in");
    ofstream fout("gym.out");
    fin>>l>>n>>m;
    if(n==m)
    {
        fout<<"1/1";
        fin.close();
        fout.close();
        return 0;
    }
    LL t=min(n,m);
    LL ans=t-1;
    if(ans>l)
    {
        fout<<"1/1";
        fin.close();
        fout.close(); 
        return 0;
    }
    LL g=__gcd(m,n);
    LL lcm=m/g*n;
    if((long double)m/g*n<=(long double)l)
        ans+=l/lcm*t;
    LL tmp=__gcd(ans,l);
    fout<<ans/tmp<<'/'<<l/tmp;
    fin.close();
    fout.close();
    return 0;
}

20180418 Mock T2——Gym

Gym