Coloring Dominoes

Coloring Dominoes

时间限制: 1 Sec 内存限制: 128 MB 
原题链接 https://arc081.contest.atcoder.jp/tasks/ARC081_B

题目描述

We have a board with a 2×N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1×2 or 2×1 square. 
Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors. 
Find the number of such ways to paint the dominoes, modulo 1000000007. 
The arrangement of the dominoes is given to you as two strings S1 and S2 in the following manner: 
Each domino is represented by a different English letter (lowercase or uppercase). 
The j-th character in Si represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

Constraints 
1≤N≤52 
|S1|=|S2|=N 
S1 and S2 consist of lowercase and uppercase English letters. 
S1 and S2 represent a valid arrangement of dominoes.

输入

Input is given from Standard Input in the following format: 
N 
S1 
S2

输出

Print the number of such ways to paint the dominoes, modulo 1000000007.

样例输入

3 
aab 
ccb

样例输出

6

题解

由于区域的宽度只有2，且每个多米诺骨牌也只能占据连续的两格，对于列而言，只有两种摆法，竖着1个或者横着两个。

#include <iostream>
#include <string>
using namespace std;
int main(){
int i,N,a;
long long ans;
string S;
string T;
cin >> N >> S >> T;;
	if(N==1){
	cout << 3 << endl; //C3 1
	}else if(N==2){
	cout << 6 << endl;  //c3 1*c2 1
	}else{
	if(S[0] == T[0]){    //竖着
		ans = 3;
		a = 1;           //判断左边是竖着还是横着 竖着a==1;横着a==0;
		i = 1;
	while(i!=N){
		if(S[i] == T[i]){
			i += 1;  //有i个竖着的，竖着的占一列
			if(a==1){
				ans = 2*ans;
				a = 1;
			}else{
				ans = ans;
				a = 1;
		}
		}else{
			i += 2; //横着的占两列
			if(a==1){ //如果左边是竖着的
				ans = 2*ans;   //和左边不一样C2 1
				a = 0;
			}else{
				ans = 3*ans;   //和左边不一样（C2 1*C1 1+C1 1*C1 1)
				a = 0;
			}
			}
		}
	}else{
		ans = 6;     //横着C3 1*C2 1
		a = 0;
		i = 2;
	while(i!=N){
		if(S[i] == T[i]){
			i +=1;
			if(a==1){
				ans = 2*ans;
				a = 1;
			}else{
				ans = ans;
				a = 1;
		}
		}else{
			i +=2;
			if(a==1){
				ans = 2*ans;
				a = 0;
			}else{
				ans = 3*ans;
				a = 0;
			}
			}
		}
	}
cout << ans%1000000007 << endl;
	}
}