zoukankan      html  css  js  c++  java
  • POJ 2039 To and Fro

    To and Fro

    Description

    Mo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is "There’s no place like home on a snowy night" and there are five columns, Mo would write down 
    t o i o y
    
    h p k n n
    e l e a i
    r a h s g
    e c o n h
    s e m o t
    n l e w x

    Note that Mo includes only letters and writes them all in lower case. In this example, Mo used the character "x" to pad the message out to make a rectangle, although he could have used any letter. 

    Mo then sends the message to Larry by writing the letters in each row, alternating left-to-right and right-to-left. So, the above would be encrypted as 

    toioynnkpheleaigshareconhtomesnlewx 

    Your job is to recover for Larry the original message (along with any extra padding letters) from the encrypted one. 

    Input

    There will be multiple input sets. Input for each set will consist of two lines. The first line will contain an integer in the range 2. . . 20 indicating the number of columns used. The next line is a string of up to 200 lower case letters. The last input set is followed by a line containing a single 0, indicating end of input.

    Output

    Each input set should generate one line of output, giving the original plaintext message, with no spaces.

    Sample Input

    5
    toioynnkpheleaigshareconhtomesnlewx
    3
    ttyohhieneesiaabss
    0

    Sample Output

    theresnoplacelikehomeonasnowynightx
    thisistheeasyoneab

    #include<iostream>
    #include<string>
    using namespace std;
    char a[400][400];
    
    int main()
    {
        string str;
        int column;
    
        while (cin >> column&&column != 0)
        {
            cin >> str;
    
            int row = str.length() / column ;
        
            for (int i = 0; i < row; i++)
                for (int j = 0; j < column; j++)
                    a[i][j] = str[i*column + ((i % 2 == 0) ? (j) : (column - j - 1))];
        
            for (int i = 0; i < column; i++)
                for (int j = 0; j < row; ++j)
                    cout << a[j][i];
        
            cout << endl;
        }
    
        return 0;
    }


  • 相关阅读:
    iOS应用崩溃日志分析
    iOS应用崩溃日志分析
    iOS 获取一个类的所有方法
    iOS 获取一个类的所有方法
    UVa 818Cutting Chains (暴力dfs+位运算+二进制法)
    UVa 1374 Power Calculus (IDA*或都打表)
    UVa 10603 Fill (暴力BFS+优先队列)
    HDU 1272 小希的迷宫 (并查集)
    HDU 1060 Leftmost Digit (数学log)
    UVa 1599 Ideal Path (两次BFS)
  • 原文地址:https://www.cnblogs.com/KennyRom/p/5831929.html
Copyright © 2011-2022 走看看