// n th number of a fibonacci series
#include<stdio.h>
#include<conio.h>
void main ()
{
int i,k = 0 , j = 1 ,n,res;
printf ( " \n enter the number " );
scanf ( "%d",&n ) ;
// printf ( "\n the series is 0, 1" ) ;
for ( i= 2 ; i<n;i++ )
{
res= k+j ;
k=j;
j=res;
// printf ( ", %d" ,res ) ;
}
printf ( " \n The n th term is %d " ,res ) ;
getch() ;
}
#include<stdio.h>
#include<conio.h>
void main()
{
int f=0,s=1,t,n,i;
clrscr();
printf("\nENTER THE VALUE OF N\n");
scanf("%d",&n);
if(n==1)
printf("FIBONACCI NUMBER IS 0");
else if(n==2)
printf("FIBONACCI NUMBER IS 1");
else
{
for(i=3;i<=n;i++)
{
t=f+s;
f=s;
s=t;
}
printf("\nFIBONACCI NUMBER IS %d",t);
}
getch();
}
What will be result of the following program?
void myalloc(char *x, int n)
{
x= (char *)malloc(n*sizeof(char));
memset(x,\0,n*sizeof(char));
}
main()
{
char *g="String";
myalloc(g,20);
strcpy(g,"Oldstring");
printf("The string is %s",g);
}
a) The string is : String
b) Run time error/Core dump
c) The string is : Oldstring
d) Syntax error during compilation
e) None of these
WAP TO ACCEPT STRING AND COUNT A COMES N TIMES B COMES N
TIMES C COMES N TIMES D COMES N TIMES AND SO ON.........
AT LAST UNTIL Z COMES N TIMES...............
Is there any restriction in how many arguments printf or
scanf function can take?
in which file in my c++ compiler i can see the code for
implementation of these two functions??