题解 | #字符串反转#

就是个斐波那契数列，当前项等于上一项和上上项之和，用递归

#include <stdio.h>


int fib(int n);

int main(void)
{
    int month;
    int sum = 0;

    while( scanf("%d", &month) != EOF)
    {
        sum = fib(month);
        printf("%d\n", sum);
    }



    return 0;
}


int fib(int n)
{
    if( (n == 1) || (n == 2) )
    {
        return 1;
    }
    else
    {
        return ( fib(n-1) + fib(n-2) );
    }
}