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;
}