In the first place, there are only \(2^3 = 8\) combinations of weather: for each of the first, second and third days, the whether is either sunny or rainy.

For example, you can get the correct answer by checking in the following order:

• check if there are \(3\) consecutive rainy days
• check if there are \(2\) consecutive rainy days
• check if there is \(1\) consecutive rainy day
• otherwise (\(0\) days)

This way, you can avoid mistakes, and moreover it is extensible to the input more than three days.

#include <iostream>
using namespace std;

int main() {
    string S;
    cin >> S;
    bool p = S[0] == 'R';
    bool q = S[1] == 'R';
    bool r = S[2] == 'R';
    if (p and q and r) {
        cout << 3 << endl;
    } else if ((p and q) or (q and r)) {
        cout << 2 << endl;
    } else if (p or q or r) {
        cout << 1 << endl;
    } else {
        cout << 0 << endl;
    }

    return 0;
}