zoukankan      html  css  js  c++  java
  • [AtCoder][ARC081]Coloring Dominoes 题解

    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: 

    S1 
    S2

    输出

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

    样例输入


    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,因此对于任意一列而言,只有两种摆法,竖着1个或者横着两个。 
    因此,通过数学规律可以计算出涂色的可能性。

    代码

    #include <iostream>
    #include <string>
    
    #define ull unsigned long long
    #define MO 1000000007
    
    using namespace std;
    int n, k, c;
    string s1, s2;
    ull ans;
    
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
    
        cin >> n >> s1 >> s2;
        if (s1[0] == s2[0])c = 1, ans = 3; else c = 2, ans = 6;
        k = c;
        while (k < n) {
            if (s1[k] == s2[k]) {
                ans *= 3 - c;
                c = 1;
            } else {
                ans *= c + 1;
                c = 2;//Skip 2
            }
            k += c;
            ans %= MO;
        }
        cout << ans;
        return 0;
    }
  • 相关阅读:
    localStorage存储数组以及取数组方法
    jq选择CheckBox进行排序
    js定时函数,定时改变字体的大小
    JQuery Datatable用法
    WebSocket实战
    代码段
    黎活明给程序员的忠告 收藏
    雅砻江后勤项目经验总结
    Java泛型方法
    回忆,梦的开始
  • 原文地址:https://www.cnblogs.com/xfl03/p/9379518.html
Copyright © 2011-2022 走看看