题解 | 牛牛学数列5

#include <stdio.h>
int fib(int n)
{
  int a = 1;
  int b = 1;//n=1;n=2时的值
  int c = 1;//解决n=1，n=2的情况
  while(n>=3)
  {
   c=a+b;
   a=b;
   b=c;
   n--;
  }
   return c;
}
int main() {
    int ret = 0;
    int n = 0;
    scanf("%d",&n);
    ret = fib(n);
    printf("%d\n",ret);
    return 0;
}