学习打卡 1-26

//循环法求斐波那契数int feibo(int n){    int a = 1;    int b = 1;    int c = 1;    while (n > 2)    {        c = a + b;        a = b;        b = c;        n--;    }    return c;}//递归法求斐波那契数（效率低下）//int feibo(int n)//{//  if (n <= 2)//  {//      return 1;//  }//  else//      return feibo(n - 1) + feibo(n - 2);//}int main(){    int n = 0;    scanf("%d", &n);    int ret = feibo(n);    printf("第%d个斐波那契数列值=%d\n",n, ret);    return 0;}

用递归求N!int Fac2(int n){    int i = 0;    int ret = 1;    for (i = 1; i <= n; i++)    {        ret *= i;    }    return ret;}//int Fac1(n)//{//  if (n == 1)//  {//      return 1;//  }//  else//  {//      return n*Fac1(n - 1);//  }//}int main(){    int n = 0;    int ret = 0;    scanf("%d", &n);    ret = Fac1(n);    printf("%d\n", ret);    return 0;}

递归初步理解int my_strlen(char* str){    if (*str != '\0')        return 1 + my_strlen(str + 1);    else        return 0;}int main(){    char arr[] = "bit";    int len = my_strlen(arr);    printf("len=%d\n",len);    return 0;}

更多相关文章

随机推荐