C语言 · 报时助手

基础练习 报时助手

时间限制：1.0s   内存限制：512.0MB

锦囊1

　　判断，字符串输出。

锦囊2

　　按要求输出，判断特殊情况。

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

代码一：

 1 /*
 2 0:zero, 1: one, 2:two, 3:three, 4:four, 5:five
 3 6:six, 7:seven, 8:eight, 9:nine, 10:ten, 11:eleven
 4 12:twelve, 13:thirteen, 14:fourteen, 15:fifteen
 5 16:sixteen, 17:seventeen, 18:eighteen, 19:nineteen
 6 20:twenty, 30:thirty，40:forty，50:fifty。
 7
 8 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。
 9     如31首先读30再加1的读法，读作“thirty one”。
10 按上面的规则:
11     21:54读作“twenty one fifty four”;
12     9:07读作“nine seven”;
13     0:15读作“zero fifteen”。
14 */
15 #include<stdio.h>
16 int main(){
17     char a[21][20]={"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen","twenty"};
18     char b[6][20]={"","","twenty","thirty","forty","fifty"};
19     int h,m;
20     scanf("%d%d",&h,&m);
21     if(m==0){
22         if(h<=20)
23             printf("%s o‘clock",a[h]);
24         else
25             printf("%s %s",b[h/10],a[h%10]);
26     }else{
27         if(h<=20){
28             if(m<=20){
29                 printf("%s %s",a[h],a[m]);
30             }else{
31                 printf("%s %s %s",a[h],b[m/10],a[m%10]);
32             }
33         }else{
34             if(m<=20){
35                 printf("%s %s %s",b[h/10],a[h%10],a[m]);
36             }else{
37                 printf("%s %s %s %s",b[h/10],a[h%10],b[m/10],a[m%10]);
38             }
39         }
40     }
41     return 0;
42 }

代码二：

 1 #include<stdio.h>
 2 void ass(int number){
 3     char n[100][100]={"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen","twenty",
 4     "thirty","forty","fifty"};
 5     int a=number/10;
 6     int b=number%10;
 7     if(number<=20){
 8         printf("%s",n[number]);
 9     }else{
10         if(b!=0){
11             printf(" %s",n[b]);
12         }
13         printf("%s",n[number+18]);
14     }
15 }
16 void time_ass(int hour,int minu){
17     if(minu==0){//整点
18         ass(hour);
19         printf(" o‘clock");
20     }else{//非整点
21         ass(hour);
22         printf(" ");
23         ass(minu);
24     }
25 }
26 int main(){
27     int h,m;
28     scanf("%d%d",&h,&m);
29     time_ass(h,m);
30     return 0;
31 }