蓝桥杯 BASIC-26 基础练习报时助手

问题描述
　　给定当前的时间，请用英文的读法将它读出来。
　　时间用时h和分m表示，在英文的读法中，读一个时间的方法是：
　　如果m为0，则将时读出来，然后加上“o’clock”，如3:00读作“three o’clock”。
　　如果m不为0，则将时读出来，然后将分读出来，如5:30读作“five thirty”。
　　时和分的读法使用的是英文数字的读法，其中0~20读作：
　　0:zero, 1: one, 2:two, 3:three, 4:four, 5:five, 6:six, 7:seven, 8:eight, 9:nine, 10:ten, 11:eleven, 12:twelve, 13:thirteen, 14:fourteen, 15:fifteen, 16:sixteen, 17:seventeen, 18:eighteen, 19:nineteen, 20:twenty。
　　30读作thirty，40读作forty，50读作fifty。
　　对于大于20小于60的数字，首先读整十的数，然后再加上个位数。如31首先读30再加1的读法，读作“thirty one”。
　　按上面的规则21:54读作“twenty one fifty four”，9:07读作“nine seven”，0:15读作“zero fifteen”。
输入格式
　　输入包含两个非负整数h和m，表示时间的时和分。非零的数字前没有前导0。h小于24，m小于60。
输出格式
　　输出时间时刻的英文。
样例输入
0 15
样例输出
zero fifteen

#include <iostream>
using namespace std;

int main() {
    int h, m;
    cin >> h >> m;
    string arr[24] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen",
					"fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty", "twenty one", "twenty two", "twenty three"};
	cout << arr[h] << " ";
    if (m == 0)
        cout << "o'clock";
    int t = m % 10;
    m = m / 10;
    switch(m) {
        case 2: cout << "twenty "; break;
        case 3: cout << "thirty "; break;
        case 4: cout << "forty "; break;
        case 5: cout << "fifty "; break;
        default: break;
    }
    if (m == 0 && t != 0) {
        cout << arr[t];
    }
    if (m == 1) {
        cout << arr[t + 10];
    }
    if (m != 0 && m != 1 && t!= 0) {
        cout << arr[t];
    }
    return 0;
}

#include <iostream>

using namespace std;

int main() {

int h, m;

cin >> h >> m;

string arr[24] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen",

"fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty", "twenty one", "twenty two", "twenty three"};

cout << arr[h] << " ";

if (m == 0)

cout << "o'clock";

int t = m % 10;

m = m / 10;

switch(m) {

case 2: cout << "twenty "; break;

case 3: cout << "thirty "; break;

case 4: cout << "forty "; break;

case 5: cout << "fifty "; break;

default: break;

}

if (m == 0 && t != 0) {

cout << arr[t];

}

if (m == 1) {

cout << arr[t + 10];

}

if (m != 0 && m != 1 && t!= 0) {

cout << arr[t];

}

return 0;

}

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