Problem:
Score : 100 points

Problem Statement

We have weather records at AtCoder Town for some consecutive three days. A string of length 3, S, represents the records - if the i-th character is S, it means it was sunny on the i-th day; if that character is R, it means it was rainy on that day.

Find the maximum number of consecutive rainy days in this period.

Constraints

|S| = 3

Each character of S is S or R.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum number of consecutive rainy days in the period.

Sample Input 1

RRS

Sample Output 1

2

We had rain on the 1-st and 2-nd days in the period. Here, the maximum number of consecutive rainy days is 2, so we should print 2.

Sample Input 2

SSS

Sample Output 2

0

It was sunny throughout the period. We had no rainy days, so we should print 0.

Sample Input 3

RSR

Sample Output 3

1

We had rain on the 1-st and 3-rd days - two "streaks" of one rainy day, so we should print 1.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int llu;
typedef long double ld;
#define vi vector<int>
#define vl vector<ll>
#define vp vector<pair<int,int>>
#define pb push_back
#define pf push_front
#define mp map<ll,ll>
#define INF 1000000000000000000
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
ll gcd(ll a,ll b)
{
    if(a==0)
    {
        return b;
    }
    else
    {
        return gcd(b%a,a);   
    }
}
ll lcm(ll a,ll b)
{
    return a*b/gcd(a,b);
}
int f(char c)
{
    return (int)c-96;
}
/**********************************************************************
***********************************************************************
*********************ACTUAL CODE BEGINS HERE***************************/
int main()
{   
    fast;
    ll x,k,d;
    cin>>x>>k>>d;
    x=abs(x);
    if(x-k*d>=0)
    {
        cout<<x-k*d;
        return 0;
    }
    ll m=INF;
    ll a=x-x/d*d;
    if((x/d)%2!=k%2)
    {
        a-=d;
    }
    for(int i=1;i<=5;i++)
    {   
        m=min(abs(a),m);
        if(a<0)
        {
            break;
        }
        a-=2*d;
    }
    cout<<m;
    return 0;
}