CodeForces 173A Rock-Paper-Scissors（数论）

题意：两个人在玩石头剪刀布，给你一个字符串表示第一个人的顺序，给你第二个字符串表示第二个人石头剪刀布的顺序，玩n局之后，问你两个人各输多少局

思路:求一个lcm之后，然后我们暴力这个lcm里面各输多少局，然后再暴力算余数里面各数多少局。

#include<bits/stdc++.h>using namespace std;string s1;string s2;int gcd(int a,int b){if (b==0)return a;return gcd(b,a%b);}int lcm(int a,int b){return a*b/gcd(a,b);}int solve(char a,char b){if (a=='R' && b == 'P')return 1;if (a=='P' && b=='S')return 1;if (a=='S' && b=='R')return 1;return 0;}int main(){int n;scanf("%d",&n);cin >> s1 >> s2;int c = lcm(s1.size(),s2.size());int ans1 = 0;int ans2 = 0;int len1 = s1.size();int len2 = s2.size();for (int i = 0;i<c;i++){ans1+=solve(s1[i%len1],s2[i%len2]);        ans2+=solve(s2[i%len2],s1[i%len1]);}ans1*=n/c;ans2*=n/c;int p = n-n/c*c;for (int i = 0;i<p;i++){ans1+=solve(s1[i%len1],s2[i%len2]);ans2+=solve(s2[i%len2],s1[i%len1]);}cout << ans1 << " " << ans2 << endl;}

Description

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

Input

7RPSRSPP

Output

3 2

Input

5RRRRRRRRR

Output

0 0

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.