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.
Answers were Sorted based on User's Feedback
Answer / ashwin murali
#include<iostream>
using namespace std;
#include<cstring>
int main(){
string str="parecisobixa";
int l;
l=str.length();
cout<<l<<endl;
for(int i=0;i<=l;i=i+2){
string sub=str.substr(i,2);
if(sub=="pa"){cout<<1; }
if(sub=="re"){cout<<2;}
if(sub=="ci"){cout<<3;}
if(sub=="vo"){cout<<4;}
if(sub=="mk"){cout<<5;}
if(sub=="xa"){cout<<6;}
if(sub=="ze"){cout<<7;}
if(sub=="bi"){cout<<8;}
if(sub=="so"){cout<<9;}
}
return 0;
}
| Is This Answer Correct ? | 7 Yes | 1 No |
#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 |
Answer / iti tomar
string [] q = new string
[10];//"no","pa","re","ci","vo","mk","xa","ze","bi","so"];
q[0] = "no";
q[1] = "pa";
q[2] = "re";
q[3] = "no";
q[4] = "ci";
q[5] = "vo";
q[6] = "mk";
q[7] = "xa";
q[8] = "ze";
q[9] = "bi";
int alen = q.Length;
//q[0] = "so";
int ln;
string result= string.Empty;
int res;
string str = (string)TextBox1.Text;
ln = str.Length;
if (ln % 2 != 0)
{
Response.Write("Invalid number");
}
else
{
string r;
for (int i = 0; i <= ln-2; i = i + 2)
{
for (int ale = 0; ale < alen; ale++)
if (str.Substring(i,2) == q[ale])
{
result = result + ale.ToString();
//i++;
//i++;
}
}
TextBox2.Text = result;
}
}
}
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / sudha
#include<conio.h>
#include<iostream>
int main()
{
char *q[] = {"no", "pa", "re", "ci", "vo", "mk", "xa",
"ze", "bi", "so"};
char *lojban, num[3];
int i,j,k;
cout << "\nEnter the value in lojban\n" ;
cin >> lojban;
i=0;
while(lojban[i++]);
num[2] = '\0';
cout << "The number is \n";
for(j=0; j<i-1; j=j+2)
{
k = 0;
strncpy(num, lojban+j, 2);
do
{
if (k == 11) { cout << "<Invalid Entry>";
break;}
}while(strcmp(num,q[k++]));
if (k != 11)
cout << k-1;
}
getch();
return(0);
}
| Is This Answer Correct ? | 1 Yes | 5 No |
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