我想在C中制作一个程序，要求用户输入一个数字，然后用英文打印该数字。
例如：
if(INPUT == 1) then print ONE
if(INPUT == 2) then print TWO
等等。它可以使用switch-case制作，如果是，但它会使代码冗长。对于少数数字它没关系，但如果我们必须写入100，那么它将是冗长的。
对此有简短的算法或想法吗？
答案 0 :(得分：10)
您可以使用下面的内容，但最多只打印数千张。我这样做是为了解决一些特定的编程问题。这就是为什么我没有超过数千人。但不难扩大数量。此外，该程序仍然可以优化或更清晰。
#include <stdio.h>
#include <string.h>
void print(int num) {
char digit [21][10] = { "", "one", "two", "three", "four", "five", "six", "seven",
"eight", "nine", "ten", "eleven", "twelve", "thirteen",
"fourteen", "fifteen", "sixteen", "seventeen", "eighteen",
"nineteen"};
char tens [11][10] = { "", "", "twenty", "thirty", "forty", "fifty", "sixty",
"seventy", "eighty", "ninety"};
char str[1000] = {0};
int prev=0, div=1000;
strcpy(str, "");
while(div) {
if ((num / div) % 10 > 0 || (div == 10 && (num%100) > 0)) {
if (prev) {
strcat(str, "and");
prev = 0;
}
switch(div) {
case 1000:
strcat(str, digit[(num / div) % 10]);
strcat(str, "thousand");
prev = 1;
break;
case 100:
strcat(str, digit[(num / div) % 10]);
strcat(str, "hundred");
prev = 1;
break;
case 10:
if ( (num%100) >= 10 && (num%100) <= 19)
strcat(str, digit[num%100]);
else {
strcat(str, tens[(num%100)/10]);
strcat(str, digit[num%10]);
}
break;
}
}
div /= 10;
}
printf("%d %s\n", num, str);
}
int main(int argc, char **argv) {
long sum = 0;
int count = 0;
if (argc <= 1) {
fprintf(stderr, "wrong number of arguments\n");
return -1;
}
print(atoi(argv[1]));
return 0;
}
答案 1 :(得分：3)
你可以使用它，它可以用来将前99个整数转换为单词。而且有点简单。看看：
void main()
{
int n,m,j;
clrscr();
printf("Enter any number between 1 to 99 : ");
scanf("%d",&n);
printf("You entered ");
if(n>0&&n<=10)
goto one;
else if (n>10&&n<20)
{
m=n%10;
goto two;
}
else if(n>20&&n<100)
{
j=n/10;
n=n%10;
goto three;
}
two:
switch(m)
{
case 1:printf("eleven ");
break;
case 2:printf("twelve ");
break;
case 3:printf("thirteen ");
break;
case 4:printf("fourteen ");
break;
case 5:printf("fifteen ");
break;
case 6:printf("sixteen ");
break;
case 7:printf("seventeen ");
break;
case 8:printf("eighteen ");
break;
case 9:printf("nineteen ");
break;
}
three:
switch(j)
{
case 2:printf("twenty ");
goto one;
case 3:printf("thirty ");
goto one;
case 4:printf("fourty ");
goto one;
case 5:printf("fifty ");
goto one;
case 6:printf("sixty ");
goto one;
case 7:printf("seventy ");
goto one;
case 8:printf("eighty ");
goto one;
case 9:printf("ninety ");
goto one;
}
one:
switch(n)
{
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;
case 10:printf("ten ");
break;
}
getch();
}
希望这有帮助。
答案 2 :(得分：3)
只需使用递归。我没有足够的时间来测试它，所以这段代码可能有些错误，但您可以轻松扩展它。
public static void convertNum(int number) {
String[] digit = { "", "one", "two", "three", "four", "five", "six",
"seven", "eight", "nine", "ten", "eleven", "twelve",
"thirteen", "fourteen", "fifteen", "sixteen", "seventeen",
"eighteen", "nineteen" };
String[] tens = { "", "", "twenty", "thirty", "forty", "fifty",
"sixty", "seventy", "eighty", "ninety" };
if (number > 0 && number < 20)
System.out.print(digit[number]);
else if (number / 1000000 > 0) {
convertNum(number / 1000000);
System.out.print(" million ");
convertNum(number % 1000000);
}
else if (number / 100000 > 0) {
convertNum(number / 100000);
System.out.print(" lukh ");
convertNum(number % 100000);
}
else if (number / 1000 > 0) {
convertNum(number / 1000);
System.out.print(" thousand ");
convertNum(number % 1000);
}
else if (number / 100 > 0) {
convertNum(number / 100);
System.out.print(" hundred ");
convertNum(number % 100);
}
else if (number / 10 >= 2) {
System.out.print(" " + tens[number / 10] + " ");
convertNum(number % 10);
}
}
convertNum (9191197);
答案 3 :(得分：1)
我很难想出一种自动化这种方法的好方法，但仍然很简短。如果你知道终点（即你想要1-100），那么你可以这样做：
char* numberArray[101] = {'Zero', 'One', 'Two' ... , 'One Hundred'};
然后当你收到输入时，只需使用该数字来访问该数组索引，它就会吐出你的答案：
int input;
cin >> input; // input = 5
cout << numberArray[input]; // outputs: Five
我道歉，如果我的语法错误，我已经做了PHP和javaScript很长时间了，我现在还不记得C语法......
答案 4 :(得分：1)
你需要的是一个递归函数，它在一个，几十，一百和几千个数字之后自我调用。
例如。
num_to_string(num = 344384)
{
if( haslakh())
num_to_string(3);print("lakh");
if( hasthou())
num_to_string(44);print("thousand");
if( hashundrer())
num_to_string(38);print("hundred");
num_to_string(4);
if( num is from 1 to 9 ) print one..nine;
if( num if from 10 to 90 ) print ten to ninty;
}
答案 5 :(得分：1)
答案 6 :(得分：0)
这是我写的，这很容易扩展到任何大小。我没有清理一些我能做的事情，但逻辑工作得非常好
import java.util.Arrays; import java.util.Scanner;
public class NumReader {
static final String[] units = {"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
static final String[] tens = {"", null, "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
static final String[] teens = {"ten", "eleven", "twelve", "thrirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
static final String hundredSuffix = "hundred";
static final String[] suffixes = {"", "thousand", "million", "billion"};
static boolean isValid(int num) {
return (num <= 1000000000 && num >= 0);
}
static String numToString(int inpNum) {
return numToString(inpNum, String.valueOf(inpNum).toCharArray());
}
static String numToString(int inpNum, char[] digits) {
return numToString(inpNum, digits, false);
}
static String numToString(int inpNum, char[] digits, boolean firstCall) {
StringBuilder b = new StringBuilder();
if (inpNum == 0 && firstCall) {
return "zero";
} else if (inpNum < 10) {
return units[inpNum];
} else if (inpNum < 20) {
return teens[inpNum - 10];
} else if (inpNum < 100) {
b.append(tens[digits[0] - '0']).append(" ").append(units[digits[1] - '0']);
return b.toString();
} else if (digits.length == 3) {
String sub = new String(Arrays.copyOfRange(digits, 1, 3));
b.append(units[digits[0] - '0']).append(" ")
.append(hundredSuffix);
sub = numToString(Integer.parseInt(sub), Arrays.copyOfRange(digits, 1, 3));
if (sub.equals("")) {
b.append(sub);
} else {
b.append(" and ").append(sub);
}
return b.toString();
} else if (digits.length > 3) {
int numSuffixes = digits.length / 3;
int initCut = digits.length % 3;
int i;
String sub, opt = "";
for (i = 0; i < numSuffixes; i++) {
int end = digits.length - 3 * i;
sub = new String(Arrays.copyOfRange(digits, end - 3, end));
sub = numToString(Integer.parseInt(sub));
opt = (sub.equals("")) ? opt : (sub + " " + suffixes[i] + " " + opt);
}
if (initCut != 0) {
sub = new String(Arrays.copyOfRange(digits, 0, initCut));
opt = numToString(Integer.parseInt(sub)) + " " + suffixes[i] + " " + opt;
}
return opt;
}
return "";
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int num = s.nextInt();
if (isValid(num)) {
System.out.println(numToString(num, String.valueOf(num).toCharArray(), true));
} else {
System.out.println("Not a valid input, num <= 1000000000");
}
}
}
答案 7 :(得分：0)
我想改善kalyan的答案。有2个错误：
像这样修改代码将解决这些错误：
void print(int num) {
char digit [21][10] = { "", "one", "two", "three", "four", "five", "six", "seven",
"eight", "nine", "ten", "eleven", "twelve", "thirteen",
"fourteen", "fifteen", "sixteen", "seventeen", "eighteen",
"nineteen"};
char tens [11][10] = { "", "", "twenty", "thirty", "forty", "fifty", "sixty",
"seventy", "eighty", "ninety"};
char str[1000] = {0};
int prev=0, div=1000;
strcpy(str, "");
while(div) {
if ((num / div) % 10 > 0 || (div == 10 && (num%100) > 0)) {
if (prev) {
strcat(str, " and");
prev = 0;
}
switch(div) {
case 1000:
if (strlen(str) > 0 && str[strlen(str) - 1] != ' ')
strcat(str, " ");
strcat(str, digit[(num / div) % 10]);
if (((num / div) % 10) > 1)
strcat(str, " thousands");
else
strcat(str, " thousand");
prev = 1;
break;
case 100:
if (strlen(str) > 0 && str[strlen(str) - 1] != ' ')
strcat(str, " ");
strcat(str, digit[(num / div) % 10]);
if (((num / div) % 10) > 1)
strcat(str, " hundreds");
else
strcat(str, " hundred");
prev = 1;
break;
case 10:
if ( (num%100) >= 10 && (num%100) <= 19)
{
if (strlen(str) > 0 && str[strlen(str) - 1] != ' ')
strcat(str, " ");
strcat(str, digit[num%100]);
}
else {
if (strlen(str) > 0 && str[strlen(str) - 1] != ' ')
strcat(str, " ");
strcat(str, tens[(num%100)/10]);
if (strlen(str) > 0 && str[strlen(str) - 1] != ' ')
strcat(str, " ");
strcat(str, digit[num%10]);
}
break;
}
}
div /= 10;
}
printf("%d %s\n", num, str);
}
答案 8 :(得分：0)
这是另一种方式。不确定优点效率与内存与速度的关系，但很容易添加代码来处理更多数字。
/* File : num_to_words_int.c
*
* Descr: Prints the equivalent number in words. '1' to 'one', etc.
* This version takes the input and converts it to a numeric vs.
* a string value. 345 vs. "345". Then uses modulus division to
* count the number of digits. The count represents the places;
* i.e. count=5 ==> 10,000 ,, 1,000 ,, 100 ,, 10 ,, 1 are the
* words that will be needed.
*
* 300 => count=3 ==>three hundred
* 30 => count=2 ==>thirtey
* 3 => count=1 ==>three
* 13 => count=2 ==>thirteen
* 3456 => count=4 ==>three thousand four hundred fifty six
*
* [345], [34 5], [3 4], [3, [0]
*
* Debugging Option:
* <run> num_to_words_int.exe number option_mask
*
* 001 - print init remainder array
* 010 - print filled remainder array
* 100 - print count, index, remainder value
*
* Author: Gene Bradshaw
* Date: 09-16-2016
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math>
void main(int argc, char *argv[])
{
const int HUNDREDS=0, THOUSANDS=1, MILLIONS=2, BILLIONS=3;
int i, count, total, remainder[12]={-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1};
long number=0, orignum=0;
char *ones[] = {"zero","one","two","three","four","five","six","seven","eight","nine"};
char *teens[] = {"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
char *tens[] = {"ten","ten","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"};
char *places[] = {"hundred","thousand","million","billion"}; // place values; 10's, 100'2, etc.
// ERRORS ARGUMENTS
if(argc < 2)
{
printf("\na number is required as input!\n\n");
exit(1);
}
else if(argc > 3)
{
printf("\nonly one number and optionally a flag are required as input!\n\n");
exit(1);
}
else printf("\n");
// CONVERT TO WORDS
if(!(number = atol(argv[1]))) // zero
{
printf("number: %d\n%s\n\n", number, ones[0]);
exit(0);
}
// Debugging
if(argv[2] && (atoi(argv[2]) & 0x01))
{
for(i=11; i>-1; i--) printf("%d %d, ", i, remainder[i]);
printf("\n\n");
}
// GET DIGITS
if(number < 0) // Remeber if negative, then make positive
{
orignum = number; number *= -1;
}
count=0;
do{
remainder[count++] = number%10;
number/=10;
}while(number); // int type var converts to '0' when # < 0
// ERROR DIGIT COUNT
if (count > 12)
{
printf("\ntoo many digits; up to 12 digits supported!\n\n");
exit(1);
}
// Debugging
if(argv[2] && (atoi(argv[2]) & 0x02))
{
for(i=11; i>-1 ; i--) printf("%d %d, ", i, remainder[i]);
printf("\n\n");
}
// DISPLAY REMAINDERS
printf("number: "); // This if for displaying the reverse remainder[].
if (orignum < 0) printf("-");
for(i=count-1; i>-1; i--)
{
if(!(i%3) && i) printf("%d,", remainder[i]);
else printf("%d", remainder[i]);
}
printf("\n");
// FIND AND PRINT WORDS
total = count; i = count-1; // counting rules
if(orignum < 0) printf("negative ");
while(count)
{
if(argv[2] && (atoi(argv[2]) & 0x04)) // Debugging
printf("\nC: %d, i: %d and R: %d\n", count, i, remainder[i]);
switch(count)
{
case 1:
// print if not teens or 0
if(remainder[i+1] != 1 && remainder[i])
printf("%s ", ones[remainder[i]]);
break;
case 2:
// teens when 2nd digit is a '1'
if(remainder[i] == 1)
printf("%s ", teens[remainder[i-1]]);
// ones when 1st digit is not a '0'
else if(remainder[i])
printf("%s ", tens[remainder[i]]);
break;
case 3: // d
if(remainder[i]){
printf("%s ", ones[remainder[i]]);
printf("%s ", places[count-3]);
}
break;
case 4: // k
if(remainder[i])
printf("%s ", ones[remainder[i]]);
if(remainder[i] || (total > 4 && total < 7))
printf("%s ", places[count-3]);
break;
// 10,000 100,000 1,000,000
// ten tho hun tho one million
case 5: // 10 k
case 8: // 10 M
case 11: // 10 B
if(remainder[i]){ printf("%s ", tens[remainder[i]]); }
break;
case 6: // 100 k
case 9: // 100 M
case 12: // 100 B
if(remainder[i]){
printf("%s ", ones[remainder[i]]);
printf("%s ", places[HUNDREDS]);
}
break;
case 7: // M
if(remainder[i])
printf("%s ", ones[remainder[i]]);
if(remainder[i] || (total > 7 && total < 10))
printf("%s ", places[MILLIONS]);
break;
case 10: // B
if(remainder[i])
printf("%s ", ones[remainder[i]]);
if(remainder[i] || (total > 10 && total < 13))
printf("%s ", places[BILLIONS]);
break;
// Add cases to increase digit count supported
//case 13: //T /*- add code here -*/ break;
default: break;
}
count--; i--;
}
printf("\n\n");
}
示例：
$>./num_to_words.exe -1000000
$>number: -1,000,000
$>negative one million
$>./num_to_words.exe 123456789011
$>number: 123,456,789,011
$>one hundred twenty three billion four hundred fifty six million seven hundred eighty nine thousand eleven
$>./num_to_words.exe 123456789012
$>number: 123,456,789,012
$>one hundred twenty three billion four hundred fifty six million seven hundred eighty nine thousand twelve
$>./num_to_words.exe -123456789012
$>number: -123,456,789,012
$>negative one hundred twenty three billion four hundred fifty six million seven hundred eighty nine thousand twelve
$>./num_to_words.exe 0
$>number: 0
$>zero
$>./num_to_words.exe 1
$>number: 1
$>one