int main(void) {
unsigned int first = 0, second = 1, n;
printf("Enter n: ");
scanf("%d", &n);
printf("Fibonacci series is: ");
while( second <= n ) {
printf("%u ", second);
first = second;
second = first + second;
}
return 0;
}
2006-08-09 21:21:16
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answer #1
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answered by swami060 3
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Yes there is a series and that series starts like this:
0, 1, 1, 2, 3, 5, 8, 13, 21 and so on...
The way Fibonacci Numbers work is that you add the two digits before the number and keep doing it, and thats how Fibonacci #s work.
2006-08-08 02:23:03
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answer #2
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answered by Daniel™ 4
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1 1 2 3 5 8 13 21 34 55
2006-08-08 02:17:55
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answer #3
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answered by rydhel1016 3
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Well this is a program where next number is some of its prevoius number
like 1,1,2,3,5,8,13....and so on
code of this program in C++ is as follows
#include
#include
using namespace std;
int main()
{
cout << "-- FIBONACCI SERIES --\n";
cout << "A program to compute the Nth value the \n"
<< "Fibonacci series.\n\n";
cout << "Enter N: ";
int N, FN;
cin >> N;
FN = static_cast(1/sqrt(5)*( pow((1+sqrt(5))/2,N) - pow((1-sqrt(5))/2,N) ));
cout << "The Nth number in the series when N=" << N
<< " is " << FN << endl;
system("PAUSE");
return 0;
}
for more help please visit link
2006-08-08 02:21:46
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answer #4
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answered by waytospark 2
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http://www.mpi-sb.mpg.de/~rybal/armc-live-termin/node2.html
2006-08-08 02:19:15
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answer #5
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answered by debashis j 2
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