C b a c d c b g a b b c
#include <stdio.h> int main (void) { puts ("a a b a b c"); return 0; }
//to generate Fibonacci series upto a range of 200....(in C).... #include<stdio.h> main() { int a,b,c,i; a=0; b=1; printf("\n FIBONACCI SERIES .....\t"); i=1; while(i<=(200-2)) { c=a+b; printf("\t%d",c); a=b; b=c; i++; } }
It means a*b = c
type in: "ABCD" in the input line.
Oh, and I mean A+B+C=BB
#include<stdio.h> #include<conio.h> void main() { printf("The first 20numbers of Fibonacci series are:"); int a=0, b=1, c, n=2; printf("%d \t, %d", &a, &b); while(n<20) { c=a+b; printf("\t %d", &c); a=b; b=c; n++; } getch(); }
g.
Bowl Championship Series
You need to be more specific in what you are asking. For example, the following meets your requirements: int main() { printf("a a b a b c\n"); return 0; }
for example if a=b+c and c>0, then a>b
b divided by 2