2. Counting in Lojban, an artificial language developed over the last fourty y

2. Counting in Lojban, an artificial language developed
over the last fourty years, is easier than in most languages
The numbers from zero to nine are:
0 no
1 pa
2 re
3 ci
4 vo
5 mk
6 xa
7 ze
8 bi
9 so
Larger numbers are created by gluing the digit togather.
For Examle 123 is pareci
Write a program that reads in a lojban string(representing
a no less than or equal to 1,000,000) and output it in
numbers.

Question Posted / ram

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2. Counting in Lojban, an artificial language developed over the last fourty years, is easier than..

Answer / ts

U can easily do this in c i am writing a code you just make its function by your self.Its easy one

#include<stdio.h>
main()
{
int i=0;
char ptr,d;
char c[20];
do
{
printf(" \n Enter the lujban String whose equivalent number is to be printed \n");
scanf("%s",c);
while(c[i]!='\0')
{

switch(c[i])
{
case'n':printf("0");
break;
case 'p':printf("1");
break;
case 'r':printf("2");
break;
case 'c':printf("3");
break;
case 'v':printf("4");
break;
case 'm':printf("5");
break;
case 'x':printf("6");
break;
case 'z':printf("7");
break;
case 'b':printf("8");
break;
case 's':printf("9");
break;
}
i++;
}
printf(" \nDo you want to continue(Y/N)");
d=getche();

}
while(d=='y'||d=='Y');

}

Is This Answer Correct ?

3 Yes

1 No

2. Counting in Lojban, an artificial language developed over the last fourty years, is easier than..

Answer / abdur rab

#include <stdio.h>

int decimal_from_logban ( char* cp_logban )
{
if ( !strcmp ( cp_logban, "no" ) ) return 0;
if ( !strcmp ( cp_logban, "pa" ) ) return 1;
if ( !strcmp ( cp_logban, "re" ) ) return 2;
if ( !strcmp ( cp_logban, "ci" ) ) return 3;
if ( !strcmp ( cp_logban, "vo" ) ) return 4;
if ( !strcmp ( cp_logban, "mk" ) ) return 5;
if ( !strcmp ( cp_logban, "xa" ) ) return 6;
if ( !strcmp ( cp_logban, "ze" ) ) return 7;
if ( !strcmp ( cp_logban, "bi" ) ) return 8;
if ( !strcmp ( cp_logban, "so" ) ) return 9;
}

void logban_2_decimal ( char* _lojban, int* _decimal )
{
char lojban_array [3];

memset ( lojban_array, '\0', 3 );
if ( ( NULL != _lojban ) && ( '\0' != *( _lojban +
2 ) ) ) {
logban_2_decimal ( _lojban + 2, _decimal +
1 );
strncpy ( lojban_array, _lojban, 2 );
*_decimal = decimal_from_logban (
lojban_array );
} else {
strncpy ( lojban_array, _lojban, 2 );
*_decimal = decimal_from_logban (
lojban_array );
}

}

int main ( int argc, char* argv [] )
{
char number_lojban [] = {"sopareci"};
int number_decimal [8];
int i = 0;

logban_2_decimal ( number_lojban, number_decimal ) ;
printf ( "\nLojban :%s", number_lojban );

printf ( "\nDecimal :" );
for ( i = 0; i < ( strlen ( number_lojban ) / 2 );
i++ )
printf ("%d", number_decimal [ i ] );

return ( 0 );
}

Is This Answer Correct ?

2 Yes

2 No

2. Counting in Lojban, an artificial language developed over the last fourty years, is easier than..

Answer / simi

In java We can do the same as
int getIntValueForString(Stirng str){
if("AB".equals(str))return 0;
.
.
.
.
.
if("rs".equals(str))return 9;
}
public static void main(Stirng [args]){
//get the String as an input from the user
//Call the method to get the int
String intValue = getIntValueForLojbanString(inputString , 0);
}
Stirng getIntValueForLojbanString(String inputString ,
String Value){
int i=0;
String intValue = new intvalue();
if(inputString != null || inputString.length()==0) return
intValue;
else
{
if(Value !=null)
intValue= Value +
getIntValueForString(inputString.chatAt[i]+inputString.chatAt[i+1]).toString();

else
intValue =
getIntValueForString(inputString.chatAt[i]+inputString.chatAt[i+1]).toString();

if(inputString.length()>2)
getIntValueForLojbanString(inputString.subString(i+3) ,
intValue);}
}
}

Is This Answer Correct ?

0 Yes

3 No

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