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 <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <math.h>
#define rep(i, a, b) for(int i = a; i < (b); ++i)
using namespace std;
typedef long long ll;

vector<pair<int, int>> to[100010];
bool visited[100010];
ll dist[100010];
ll gcd(ll a, ll b)
{
   if (b == 0)
   {
       return a;
   }
    return gcd(b, a%b);
}
void dfs(int now, ll d) {
  visited[now] = true;
  dist[now] = d;
  for (auto p : to[now]) {
    if (visited[p.first]) continue;
    dfs(p.first, d + p.second);
  }
}
vector<int>ans(10005,0);
void f(ll n){
    for (ll x = 1; x*x<n;x++){
        for (ll y = 1; y*y<n;y++){
            for (ll z = 1; z*z<n;z++){
                ll v = (x*x)+(y*y)+(z*z)+(x*y)+(y*z)+(z*x);
                if(v<10005) ans[v]++;
            }
        }
    }
}
int find(ll k) {
    ll tag = 7777777;
    int cnt = 0;
    cout << tag%k << endl;
    while (tag%k) {
        cnt++;
        tag/=10;
        cout << tag << endl;
        
    }
    return cnt;
}

int main() {
    int a,b;
    string s;
    cin >> s;
    ll ans = 0;
    ll sum = 0;
    char now = 'N';
    rep(i,0,s.size()){
        if(now != 'N' && now=='R'){
            ans++;
        } else {
            ans = 0;
        }
        now=s[i];
        sum = max(ans,sum);
    }
    cout << sum << endl;
    return 0;
}