o.boj 1081 Bovine Latin-优快云博客

本文链接：https://blog.youkuaiyun.com/wenlele/article/details/7079489

本文介绍了一个将英文单词转换为Bovine Latin的简单程序。该程序接收一系列英文单词作为输入，并根据特定规则将其转化为Bovine Latin形式。文章包含完整的C++代码实现。

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注：最近这一系列ACM的内容，都是2年多之前的代码，自己回顾一下。

Bovine Latin

Submit: 947 Accepted:514

Time Limit: 1000MS Memory Limit: 65536K

Description

The cows have heard that the pigs use a secret language called "Pig Latin" when they want to communicate with each other without Farmer John being able to understand what they are saying. Thinking this is an excellent idea, they have invented their own version, aptly called Bovine Latin.

Converting an English word to a Bovine Latin word is quite simple. For words that start with a vowel ('a', 'e', 'i', 'o' or 'u'), "cow" is added to the end of the word; for example, "udder" becomes "uddercow". For words that do not begin with a vowel, the first letter is moved to the end of the word, and "ow" is added; e.g., "farmer" becomes "armerfow". So "the cows escape at dawn" becomes "hetow owscow escapecow atcow awndow." The cows fervently believe that FJ will not understand this subterfuge.

Never known as enthusiastic linguists, the cows find this translation quite tedious and thus have asked you to write a program that will take single words and translate them into Bovine Latin. They will provide you with N (1 <= N <= 100) words to translate; word lengths range from 3 to 40 letters.

Input

* Line 1: A single integer: N

* Lines 2..N+1: One word per line.

Output

* Lines 1..N: The Bovine Latin translations of the given words

Sample Input

udder

farmer

milk

aaa

zzz

Sample Output

uddercow

armerfow

ilkmow

aaacow

zzzow

Source

USACO MAR07 BRONZE Division

简单的模拟题

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string S[105];
    int N;
    char ch;
    
    cin >> N;
    
    for (int i = 0; i < N; i++)
        cin >> S[i];
    
    for (int i = 0; i < N; i++)
    {
        if (S[i][0] == 'a' || S[i][0] == 'e' || S[i][0] == 'i' ||
           S[i][0] == 'o' || S[i][0] == 'u' )
            cout << S[i] << "cow" << endl;
        else
        {
            ch = S[i][0];
            S[i].replace(0, 1, "");
            S[i].insert(S[i].end(), ch);
            cout << S[i] << "ow" << endl;       
        }
    }
    // system("pause");
    return 0;
}