CodeForces 173A

 Rock-Paper-Scissors

Nikephoros and Polycarpus play rock-paper-scissors. The loser gets pinched (not too severely!).

Let us remind you the rules of this game. Rock-paper-scissors is played by two players. In each round the players choose one of three items independently from each other. They show the items with their hands: a rock, scissors or paper. The winner is determined by the following rules: the rock beats the scissors, the scissors beat the paper and the paper beats the rock. If the players choose the same item, the round finishes with a draw.

Nikephoros and Polycarpus have played n rounds. In each round the winner gave the loser a friendly pinch and the loser ended up with a fresh and new red spot on his body. If the round finished in a draw, the players did nothing and just played on.

Nikephoros turned out to have worked out the following strategy: before the game began, he chose some sequence of items A = (a₁, a₂, ..., a_m), and then he cyclically showed the items from this sequence, starting from the first one. Cyclically means that Nikephoros shows signs in the following order: a₁, a₂, ..., a_m, a₁, a₂, ..., a_m, a₁, ... and so on. Polycarpus had a similar strategy, only he had his own sequence of items B = (b₁, b₂, ..., b_k).

Determine the number of red spots on both players after they've played n rounds of the game. You can consider that when the game began, the boys had no red spots on them.

Input

The first line contains integer n (1 ≤ n ≤ 2·10⁹) — the number of the game's rounds.

The second line contains sequence A as a string of m characters and the third line contains sequence B as a string of k characters (1 ≤ m, k ≤ 1000). The given lines only contain characters "R", "S" and "P". Character "R" stands for the rock, character "S" represents the scissors and "P" represents the paper.

Output

Print two space-separated integers: the numbers of red spots Nikephoros and Polycarpus have.

Sample Input

7
RPS
RSPP

Sample Output

3 2

Hint

In the first sample the game went like this:

• R - R. Draw.
• P - S. Nikephoros loses.
• S - P. Polycarpus loses.
• R - P. Nikephoros loses.
• P - R. Polycarpus loses.
• S - S. Draw.
• R - P. Nikephoros loses.

Thus, in total Nikephoros has 3 losses (and 3 red spots), and Polycarpus only has 2.

题意：

给你两个人石头剪刀布的出手顺序，且按照这个顺序玩n轮，问你 最后 第一个人输了几把，第二个人输了几把

题解：

求出长度的lcm ，就好做了

#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
const int N=5500;
const int inf = 899999;


int check(char a,char b) {
    if(a == b) return 0;
    else if(a == 'R' && b == 'S') return 1;
    else if(a == 'S' && b == 'P') return 1;
    else if(a == 'P' && b == 'R') return 1;
    else return -1;
}
char a[N],b[N];
int h[1001000],g[1001000];
int main() {
    int n;
    scanf("%d%s%s",&n,a,b);
    int l = strlen(a), r = strlen(b);
    int d = __gcd(l,r);
    int lcm = l * r / d;
    int  cnt = 0, tmp = 0;
    ll ans = 0 , sum = 0;
    for(int i = 1; i <= lcm; i++) {
        int now = check(a[cnt],b[tmp]);
        if(now == 1) ans++;
        else if(now == -1) sum++;
        tmp++,cnt++;
        if(tmp == r) tmp = 0;
        if(cnt == l) cnt = 0;
        h[i] = ans;
        g[i] = sum;
    }
    sum *= (n/lcm);
    ans *= (n/lcm);
    sum += g[n%lcm];
    ans += h[n%lcm];
    printf("%I64d %I64d
",sum,ans);
    return 0;
}