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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.

Answer Posted / manoj pathak

#include<iostream.h>
#include<conio.h>

void main()
{
int i=0,k;
char ph,ch[10];
clrscr();
while(ph!='\r')
{
ph=getche();
ch[i]=ph;
i++;
}
cout<<endl<<endl<<endl;
for(k=0;k<i;)
{
if(ch[k]=='n')
{
if(ch[k+1]=='o')
cout<<"0";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}
if(ch[k]=='p')
{
if(ch[k+1]=='a')
cout<<"1";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='r')
{
if(ch[k+1]=='e')
cout<<"2";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='c')
{
if(ch[k+1]=='i')
cout<<"3";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='v')
{
if(ch[k+1]=='o')
cout<<"4";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='m')
{
if(ch[k+1]=='k')
cout<<"5";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='x')
{
if(ch[k+1]=='a')
cout<<"6";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='z')
{
if(ch[k+1]=='e')
cout<<"7";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='b')
{
if(ch[k+1]=='i')
cout<<"8";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

if(ch[k]=='s')
{
if(ch[k+1]=='o')
cout<<"9";
else
{
clrscr();
cout<<"Entered String is not code" ;
break;
}
}

k=k+2;
}
getch();
}

Is This Answer Correct ?    4 Yes 4 No



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