Rock_Paper_Scissors in C

editorial
1

Explanation:
To win the game, Chef needs to score more than ⌊N/2⌋ (i.e. W = N/2+1) points. It doesn’t matter if Chef loses or draws D = N - (N/2+1) rounds.

Among the characters ‘R’, ‘P’ and ‘S’, ‘P’ is lexicographically smallest.
To find the lexicographically smallest sequence of moves, Chef will play ‘P’ move D times from the beginning of the game. Each time when Chef loses or draws playing move ‘P’, D will be decreased by 1. When D becomes 0 proceeding in this fashion, then Chef will play the winning moves (i.e. ‘R’ against ‘S’, ‘P’ against ‘R’ and ‘S’ against ‘P’).

Code:

#include <stdio.h>

int main() {
    int t;
    scanf("%d", &t);

    while (t--) {
        int n;
        scanf("%d", &n);

        char s[n + 1];
        scanf("%s", s);

        char ans[n + 1];

        for (int i = 0; i < n; i++) {
            if (s[i] == 'R') {
                ans[i] = 'P';
            } else if (s[i] == 'P') {
                ans[i] = 'S';
            } else {
                ans[i] = 'R';
            }
        }
        ans[n] = '\0';

        int rep = (n - 1) / 2;

        for (int i = 0; i < n; i++) {
            if (rep > 0 && ans[i] != 'P') {
                rep--;
                ans[i] = 'P';
            }
        }

        printf("%s\n", ans);
    }

    return 0;
}

Hope it helps! :slight_smile: