编写递归函数求解斐波那契数列其公式为Fn= Fn-1+Fn-2（n>2），其中F₁=F₂=1并分析递归调用过程

#include

using namespace std;

int fib(int n);

int main()

{

int n, answer;

cout <<"Enter number: ";

cin >> n;

cout <<"\n\n";

answer = fib(n);

cout << answer <<" is the "<< n <<"th Fibonacci number\n";

return 0;

}

int fib (int n)

{

cout <<"Processing fib("<< n <<")... ";

if (n < 3 )

{

cout <<"Return 1!\n";

return (1);

}

else

{

cout <<"Call fib("<< n-2 <<") and fib("<< n-1 <<").\n";

return( fib(n-2) + fib(n-1));

}

}