解题思路:
x/21: 来表示20以内的数;超过部分用求余 x%21 来表示; 用得到的数的大小表示 a[i]中的 i 的值,从而表示出字符串 char a[21][10]={"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven",
"twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen","twenty"};
注意事项:
分界点 20;与带零的部分是否需要输出zero
参考代码:
#include<stdio.h>
int main()
{
int h,m,h1,m1,h2,m2;
char a[21][10]={"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven",
"twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen","twenty"};
//0_20的数字用二维数组表示
scanf("%d%d",&h,&m);
h1=h/21; //根据h1得出是否可以用数组a
h2=h%21; //h2求出个位数上的数组
m1=m/10;
m2=m%10; //m同理
if(m==0)
{
if(h1==0)
{
printf("%s ",a[h]);
}
else
{
printf("%s ",a[h-h2-1]);
printf("%s ",a[h2+1]);
}
printf("o'clock"); // m=0, o'clock
}
else //m!=0;
{
if(h1==0)
{
printf("%s ",a[h]);
}
else
{
printf("%s ",a[h-h2-1]);
printf("%s ",a[h2+1]);
}
if(m1==0||m1==1) // 是否可以用数组a
{
printf("%s ",a[m]);
}
else
{
if(m1==2) // m>20; 分别输出
{
printf("twenty ");
}
else if(m1==3)
printf("thirty ");
else if(m1==4)
{
printf("forty ");
}
else if(m1==5)
{
printf("fifty ");
}
else
{
printf("sixty ");
}
if(m2!=0)
{
printf("%s ",a[m2]); // m==整十数,没有zero
}
}
}
return 0;
}
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