如何用C语言将输入的数字转化成英语

Posted 2023-03-16

数字为1-10000之间，由于我刚开始学C，我用的一种比较传统的方法比较复杂要写很多语句，所以想请教一下高手有没有什么简单的方法，谢谢！
是将数字转化成英文啊 如；11：ELEVEN，135：ONE HUNDRED FIVE

对照字母的ASCII码进行输入数字，然后再输出字母。

和你的相比，不知是否复杂。
此程序的计算范围：0<=num<1000。如果还想要计算更大的数，可以在最后面加判断语句，方法类似。
#include<stdio.h>
void main()

char *Eng1[20]="zero","one","two","three","four","five","six","seven",
"eight","nine","ten","eleven","twelve","thirteen","fourteen","fifteen",
"sixteen","seventeen","eighteen","nineteen";
char *Eng2[8]="twenty","thirty","fourty","fifty","sixty","seventy","eighty","ninety";
int num;
printf("请输入数字: ");
scanf("%d",&num);
printf("对应的英文为: ");
if(num>=0&&num<=19)
printf("%s\n",Eng1[num]);
else if(num<100)

int s,y;
s=num/10;
y=num%10;
printf("%s %s\n",Eng2[s-2],Eng1[y]);

else if(num<1000)

int b,s,y;
b=num/100;
y=num%100;
if(y>9)

s=(num%100)/10;
y=(num%100)%10;
if(y==0)
printf("%s hundred and %s\n",Eng1[b],Eng2[s-2]);
else
printf("%s hundred and %s %s\n",Eng1[b],Eng2[s-2],Eng1[y]);

else
printf("%s hundred and %s\n",Eng1[b],Eng1[y]);

参考技术A 只有按照ASCII码进行对应转换，把输入的数字转换为相应的英文字母，没有其他方法了 参考技术B #include <stdio.h>
#include <stdlib.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int main(int argc, char *argv[])
char a;
while('\n'!=a)
a=getchar();
switch (a)

case '0':printf("zero ");break;
case '1':printf("one ");break;
case '2':printf("two ");break;
case '3':printf("three ");break;
case '4':printf("four ");break;
case '5':printf("five ");break;
case '6':printf("six ");break;
case '7':printf("seven ");break;
case '8':printf("eight ");break;
case '9':printf("nine ");break;


return 0;
参考技术C #include<stdio.h>
char *Eng1[20] = "zero", "one", "two", "three", "four", "five", "six", "seven",
"eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",
"sixteen", "seventeen", "eighteen", "nineteen" ;
char *Eng2[8] = "twenty", "thirty", "fourty", "fifty", "sixty", "seventy", "eighty", "ninety" ;
int num;
void fun(int num)

if (num >= 0 && num <= 19)
printf("%s\n", Eng1[num]);
else if (num<100)

int s, y;
s = num / 10;
y = num % 10;
printf("%s %s\n", Eng2[s - 2], Eng1[y]);

else if (num<1000)

int b, s, y;
b = num / 100;
y = num % 100;
if (y>9)

s = (num % 100) / 10;
y = (num % 100) % 10;
if (y == 0)
printf("%s hundred %s ", Eng1[b], Eng2[s - 2]);
else
printf("%s hundred %s %s", Eng1[b], Eng2[s - 2], Eng1[y]);

else
printf("%s hundred %s", Eng1[b], Eng1[y]);


void main()

scanf("%d", &num);
if (num > 0)

if (num > 1000)

if (num > 1000000)

if (num > 1000000000)

int numw = num / 1000000000;
num = num % 1000000000;
fun(numw);
printf(" billion ");

int numx = num / 1000000;
num = num % 1000000;
fun(numx);
printf(" million ");

int numy = num / 1000;
num = num % 1000;
fun(numy);
printf(" thousand ");

int numz = num;
fun(num);


根据我的题目中间不用and，要改可以自调，数据范围一直到100billon，在此谢谢mhy8946（不会是清洁工吧。。。同行么）的算法，要不然我也搞不出来hh 参考技术D 没有

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