蓝桥杯基础练习VIP-报时助手-题解（C++代码）利用开关语句，一个函数解决。

解题思路:

条件明确，数字和字母表示一一对应。所以想到运用开关语句。

注意事项:

1.函数封装。为了减少操作，利用递归思想对20到60间的非特殊数字采用递归。

参考代码:

#include<iostream>

#include<string>

using namespace std;

string s(int x) {

string y;

switch (x) {

case 0: y = "zero";

break;

case 1:y = "one";

break;

case 2:y = "two";

break;

case 3:y = "three";

break;

case 4:y = "four";

break;

case 5:y = "five";

break;

case 6:y = "six";

break;

case 7:y = "seven";

break;

case 8:y = "eight";

break;

case 9:y = "nine";

break;

case 10:y = "ten";

break;

case 11:y = "eleven";

break;

case 12:y = "twelve";

break;

case 13:y = "thirteen";

break;

case 14:y = "fourteen";

break;

case 15:y = "fifteen";

break;

case 16:y = "sixteen";

break;

case 17:y = "seventeen";

break;

case 18:y = "eighteen";

break;

case 19:y = "nineteen";

break;

case 20:y = "twenty";

break;

case 30:y = "thirty";

break;

case 40:y = "forty";

break;

case 50:y = "fifty";

break;

default:

y = s(x - x % 10) +" "+ s(x % 10);  //递归一次，代码复用

break;

}

return y;

}

int main(){

int h, m;

cin >> h >> m;

string x, y;

if (m != 0) {

x = s(h);

y = s(m);

cout << x << " " << y << endl;

}

else {//m==0

x = s(h);

cout << x << " o'clock" << endl;

}

return 0;

}