Write a program that accepts a number and output its equivalent in words?

Answer 1

/*mycfiles.wordpress.com
program to convert digits to character*/
#include
#include
void main()
{
int a[5],i,n;
clrscr();
cout<<"Enter the Value";
cin>>n;
for(i=4;i>=0;i--)
{
a[i]=n%10;
n=n/10;
}
for(i=0;i<5;i++)
{
if(a[i]!=0)
{
switch(a[i])
{
case 0:cout<<"Zero";
break;
case 1:cout<<"One";
break;
case 2:cout<<"Two";
break;
case 3:cout<<"Three";
break;
case 4:cout<<"Four";
break;
case 5:cout<<"Five";
break;
case 6:cout<<"Six";
break;
case 7:cout<<"Seven";
break;
case 8:cout<<"Eight";
break;
case 9:cout<<"Nine";
break;
}
}
}
getch();
}

------------------------------------------------------------------------------------

Program to Convert Numbers into Words

#include

void pw(long,char[]);
char *one[]={" "," one"," two"," three"," four"," five"," six"," seven","
eight"," Nine"," ten"," eleven"," twelve"," thirteen"," fourteen","
fifteen"," sixteen"," seventeen"," eighteen"," nineteen"};
char *ten[]={" "," "," twenty"," thirty"," forty"," fifty"," sixty","
seventy"," eighty"," ninety"};


void main()
{
long n;
clrscr();
printf("
Enter any 9 digit no: ");
scanf("%9ld",&n);
if(n<=0)
printf("Enter numbers greater than 0");
else
{
pw((n/10000000),"crore");
pw(((n/100000)%100),"lakh");
pw(((n/1000)%100),"thousand");
pw(((n/100)%10),"hundred");
pw((n%100)," ");
}
getch();
}


void pw(long n,char ch[])
{
(n>19)?printf("%s %s ",ten[n/10],one[n%10]):printf("%s ",one[n]);
if(n)printf("%s ",ch);
}


// for any query visit
// "http://www.c.happycodings.com/Beginners_Lab_Assignments/code51.html"

Program to Convert Numbers into Words

#include

void pw(long,char[]);
char *one[]={" "," one"," two"," three"," four"," five"," six"," seven","
eight"," Nine"," ten"," eleven"," twelve"," thirteen"," fourteen","
fifteen"," sixteen"," seventeen"," eighteen"," nineteen"};
char *ten[]={" "," "," twenty"," thirty"," forty"," fifty"," sixty","
seventy"," eighty"," ninety"};


void main()
{
long n;
clrscr();
printf("
Enter any 9 digit no: ");
scanf("%9ld",&n);
if(n<=0)
printf("Enter numbers greater than 0");
else
{
pw((n/10000000),"crore");
pw(((n/100000)%100),"lakh");
pw(((n/1000)%100),"thousand");
pw(((n/100)%10),"hundred");
pw((n%100)," ");
}
getch();
}


void pw(long n,char ch[])
{
(n>19)?printf("%s %s ",ten[n/10],one[n%10]):printf("%s ",one[n]);
if(n)printf("%s ",ch);
}


// for any query visit
// "http://www.c.happycodings.com/Beginners_Lab_Assignments/code51.html"

Answer 2

/**

* Note: this should work for all 0 <= num < 1,000,000,000

* @return a String representing the English-word value of num

*/

public static final String numberToWord(final int num) {

// special case to simplify later on

if (num == 0) {

return "zero";

}

// constant number names for each category

// single digits

final String n_1_9[] = new String[]{"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

// unfortunate special cases for ten, eleven, twelve, and teens

final String n_10_19[] = new String[]{"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};

// tens

final String n_20_90[] = new String[]{"", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};

final String n_100 = "hundred";

final String n_1000 = "thousand";

final String n_1000000 = "million";

// use StringBuilder for efficient modifications

StringBuilder numWord = new StringBuilder();

int n = num;

// append with selective recursion for all our cases

if (n >= 1000000) {

numWord.append(numberToWord(n / 1000000));

numWord.append(' ');

numWord.append(n_1000000);

numWord.append(' ');

n %= 1000000;

}

if (n >= 1000) {

numWord.append(numberToWord(n / 1000));

numWord.append(' ');

numWord.append(n_1000);

numWord.append(' ');

n %= 1000;

}

if (n >= 100) {

numWord.append(n_1_9[n / 100]);

numWord.append(' ');

numWord.append(n_100);

numWord.append(' ');

n %= 100;

}

if (n >= 20) {

numWord.append(n_20_90[(n / 10) - 1]);

numWord.append(' ');

n %= 10;

}

if (n >= 10) {

numWord.append(n_10_19[n - 10]);

}

if (n < 10) {

numWord.append(n_1_9[n]);

}

return numWord.toString().trim();

}

Answer 3

//Program to Print a Number into Word (Between 1 to 9)

#include
#include
void main()
{
int n;
clrscr();
printf("Enter number between 1-9:");
scanf("%d",&n);
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;
default:
printf("Contains More than 1 Digit");
}
getch();
}

output:
Enter number between 1-9:
2
Two
Enter number between 1-9:
230
Contains More than 1 Digit

Answer 4

All you need to do is explain to your teacher how it works.

public class Main {

public static void main(String[] av) { System.out.println(cvt(12345)); }

static String ten[] = { "", " ten", " twenty", " thirty", " fourty",

" fifty", " sixty", " seventy", " eighty", " ninety" };

static String one[] = { "", " one", " two", " three", " four",

" five", " six", " seven", " eight", " nine",

" ten", " eleven", " twelve", " thirteen", " fourteen",

" fifteen", " sixteen", " seventeen", " eighteen", " nineteen" };

static String cvt(int a) { return cvt5(a).substring(1);}

static String cvt5(int a) { return (a >= 1000 ? cvt3(a / 1000) + " thousand" : "") + cvt3(a % 1000); }

static String cvt3(int a) { return (a >= 100 ? cvt2((a / 100)%10) + " hundred" : "") + cvt2(a % 100); }

static String cvt2(int a) { return a < 20 ? one[a] : (ten[a / 10] + one[a % 10]); }

}

Answer 5

class abc

{

void main( int num )

{

int d = 1;

int dig = 0;

int temp = num;

System.out.println(temp);

while(num > 9)

{

num = num / 10 ;

d = d * 10;

}

while( d != 0)

{

dig = temp / d;

if( dig 0)

System.out.print("Zero");

temp = temp - (dig * d);

d = d / 10 ;

}

}

}

Answer 6

Your question is inconclusive. Please provide an example of what you are trying to achieve

Answer 7

Write a program enters the ten-digit number and prints

Using the largest for

Answer 8

write a java program to implement Dynamic method dispatch