CodeForces - 1260C (+ greedy thinking)

The meaning of problems

https://vjudge.net/problem/CodeForces-1260C

A string of bricks, usually a multiple of r multiples of b must be not painted red, all multiples of b rather than a multiple of r must be painted blue, are selected from a common multiple of the coating. After the painted brick elected, and asked whether there must be a continuous k-brick the same color.

Thinking

When r and b have a common factor (i.e. gcd! = 1), it can be found just skips some number, and no difference in its simplest form. So let's reduction, the reduction, if r == b, then definitely the solvability of (post questions Tu); otherwise, assume b> r, then the sequence must be similar xrbrbrrb, because faster growth b, then for a continuous period of r, we use two b to separate them, the number of positions between the two and b is b-1, k r to be occupied by a length of (k-1) * r + 1, is determined two b between the two can not be used to fill the k r, if or how the position, then not; otherwise feasible.

Code

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
const int N=200005;
const int mod=1e9+7;
const double eps=1e-8;
const double PI = acos(-1.0);
#define lowbit(x) (x&(-x))
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
int main ()
{
    std::ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--)
    {
        ll r,b,k;
        cin>>r>>b>>k;
        if(r>b)
            swap(r,b);
        ll g=gcd(r,b);
        r/=g,b/=g;
        if(r==b)
        {
            cout<<"OBEY"<<endl;
        }
        else if(b-1<(k-1)*r+1)
            cout<<"OBEY"<<endl;
        else
            cout<<"REBEL"<<endl;
    }
    return 0;
}