算法取一個數字並輸出其英文單詞

Question

我想用C語言編寫一個程序，它會要求用戶輸入一個數字，然後用英文輸出該數字。

例如：

if(INPUT == 1) then print ONE 
if(INPUT == 2) then print TWO 

等。它可以使用開關大小寫，否則它會使代碼冗長。對於少數數字，這很好，但如果我們必須寫入100，那麼它會很長。

是否有一個簡短的算法或想法呢？

Answer 1

您可以使用下面的內容，但這隻能打印多達數千個。我這樣做是爲了解決一些特定的編程問題。這就是爲什麼我沒有超過數千人。但不難擴展到更大的數量。而且，這個程序還可以進行優化或更清晰。

#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; 
} 

Answer 2

你可以用它來將最多99個整數轉換成單詞。它有點簡單。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(); 
} 

希望這會有所幫助。

Answer 3

我很難想出一個自動執行此操作的好方法，並且仍然使其縮短。如果你知道終點（即你想去1-100），那麼你可以做這樣的事情：

char* numberArray[101] = {'Zero', 'One', 'Two' ... , 'One Hundred'}; 

然後當你接收輸入，簡單地用數字來訪問數組索引，和它會吐出你的答案：

int input; 
cin >> input; // input = 5 
cout << numberArray[input]; // outputs: Five 

我道歉，如果我的語法是錯誤的，我一直在做PHP和JavaScript了這麼久，現在我不記得C語法那麼好......

Answer 4

你需要的是一個遞歸函數，它自己調用自己的數字，數十，數字和千位數。

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; 
} 

Answer 5

只要使用遞歸。我沒有足夠的時間來測試它，所以這段代碼可能是錯誤的，但你可以很容易地擴展它。

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); 

Answer 6

這就是我寫的，這很容易擴展到任何大小。我還沒有清理一些我可以但邏輯工作很好的東西

import java.util.Arrays; import java.util.Scanner;

公共類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"); 
    } 
} 

}

Answer 7

我會感到驚訝，如果人類沒有實現它千（抱歉：1000）已。

So I checked on Github and a found at least a few。

我沒有貢獻任何一個。

Answer 8

我想改善kalyan的答案。有2個錯誤：

• 單詞之間沒有空格。例如：100將得到結果one hundred（無空格）
• 在英語中，一百和一千個複數形式都需要s。例如200 - >兩百（它需要的「在結果）

這樣的修改代碼將解決這些錯誤：

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); 

} 

Answer 9

這裏的另一種方式。不確定優點效率與內存與速度的關係，但很容易添加代碼來處理更多數字。

/* 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"); 
} 

例子：

$>./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