CodeForces 173A Rock-Paper-Scissors 数学

Rock-Paper-Scissors

题目连接：

http://codeforces.com/problemset/problem/173/A

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, ..., am), 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, ..., am, a1, a2, ..., am, a1, ... and so on. Polycarpus had a similar strategy, only he had his own sequence of items B = (b1, b2, ..., bk).

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

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

然后玩n局之后，问你两个人各输多少局

题解：

就求一个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。

就好了

这样复杂度可以降为O(lcm)的

代码

#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,ans2=0;
    int len1 = s1.size(),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;
}