C语言 · 报时助手

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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         else25             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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