Seek to achieve by using the n-th function Fibonacei / k is calculated to achieve n ^ / obtaining a non-negative integers the sum of its digital / character string parameter oppositely oriented / strlen implemented factorial / a n / every print integers

1. The non-recursive and recursive are seeking to achieve the n-th Fibonacci number.

Using recursive function to implement, the disadvantage is inefficient, from scratch every time call
example:
// 50
// 49 48
// 48474746

#include<stdlib.h>
#include<stdio.h>
//1 1 2 3 5 8 13  21 34 55......
int fib(int n)//利用递归函数求取
{
	if (n <= 2)
		return 1;
	else
		return fib(n - 1) + fib(n - 2);
}
int main()
{
	int n = 0;
	int ret = 0;
	printf("求第n个斐波那契数：\n");
	scanf("%d", &n);
	ret=fib(n);
	printf("%d\n", ret);
	system("pause");
	return 0;
}

Definition of a, b, c, a number of each calculation, a backward movement of the abc, avoiding double counting

#include<stdlib.h>
#include<stdio.h>
//1 1 2 3 5 8 13 21 34 55
//a b c
int fib(int n)
{
	int a = 1;
	int b = 1;
	int c = 1;
	while (n > 2)
	{
		c = a + b;
		a = b;
		b = c;
		n--;
	}
	return c;
}

2. Write a function to achieve the calculated n ^ k

#include<stdlib.h>
#include<stdio.h>
int Index(int n, int k)
{
	if (k == 1)
		return n;
	else
		return n*Index(n, k - 1);
}
int main()
{
	int n;
	int k;
	int ret;
	printf("请分别输入数字n和指数k\n");
	scanf_s("%d,%d", &n, &k);
	ret= Index(n, k);
	printf("ret=%d\n", ret);
	system("pause");
	return 0;
}

3. Write a recursive function DigitSum (n), enter a non-negative integer, it returns the sum of the numbers and the composition,
   e.g., call DigitSum (1729), it should return 1 + 2 + 7 + 9, and it is 19

#include<stdlib.h>
#include<stdio.h>
void DigitSum(int n,int* sum)
{
	if (n > 9)
	{
		DigitSum(n / 10, sum);
	}
	*sum += n % 10;
}
//1729
//(172) 9
//(17) 2 9
//1 7 2 9
int main()
{
	int num;
	int sum=0;
	printf("请输入一个非负整数：\n");
	scanf("%d", &num);
	DigitSum(num, &sum);
	printf("%d", sum);
	system("pause");
	return 0;
}

4. Write a function reverse_string (char * string) (recursive)
   implementation: the parameter string of characters oppositely oriented.
   Requirements: You can not use string manipulation functions in the C library.

#include<stdlib.h>
#include<stdio.h>

void  reverse_string(char * string)
{
	if (*(string) != '\0')
	{
		reverse_string(string + 1);
	}
	printf("%c", *(string-1));
}
int main()
{
	char arr[] = "abcd";
	reverse_string(arr);
	printf("\n");
	system("pause");
	return 0;
}

The non-recursive and recursive were achieved strlen

#include<stdio.h>
#include<stdlib.h>
int Strlen(char* arr)//递归的方式实现
{
	if (*(arr) != '\0')
	{
		return Strlen(++arr)+1;
	}
	return 0;
}
int len(char* arr)//非递归的方式
{
	int count = 0;
	while (*(arr + count) != '\0')
	{
		count++;
	}
	return count;
}
int main()
{
	int ret;
	char arr[] = "abcd";
	ret=Strlen(arr);
	printf("%d\n", ret);
	ret = len(arr);
	printf("%d\n", ret);
	system("pause");
	return 0;
}

6. The non-recursive and recursive are seeking to achieve the factorial of n

#include<stdio.h>
#include<stdlib.h>
int Fac(int i)//用递归函数实现n的阶乘
{
	if (i <= 1)
	{
		return 1;
	}
	return i*Fac(i - 1);
}
int fac(int i)//用非递归函数实现
{
	int n = 1;
	int ret = 1;
	for (n = 1; n <= i; ++n)
	{
		ret = ret*n;
	}
	return ret;
}
int main()
{
	int i;
	printf("请输入想要所求数字n的阶乘：\n");
	scanf("%d", &i);
	Fac(i);
	printf("%d\n", Fac(i));
	fac(i);
	printf("%d\n", fac(i));
	system("pause");
	return 0;
}

7. recursive manner every print an integer

#include<stdio.h>
#include<stdlib.h>
int Print(int n)
{
	if (n > 9)
	{
		Print(n/10);
	}
	printf("%d ", n % 10);
}
int main()
{
	int i;
	printf("请输入一个整数:\n");
	scanf_s("%d", &i);
	Print(i);
	system("pause");
	return 0;
}