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  • 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 }
     
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  • 原文地址:https://www.cnblogs.com/panweiwei/p/6480899.html
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