本文共 2022 字，大约阅读时间需要 6 分钟。

1、1 到 n 的累加

#include
   
    void main(){   	int i;	int n;	scanf("%d",&n);	int sum( int n);	printf("%d",sum(n));}int sum(int n){   	if(n < 0)		return 0;	if(n == 1)		return 1;	else		return n + sum(n-1);}
   

2、递归求阶乘

#include
   
    double fact(int n);double factsum(int n);double main(){   	int n;	scanf("%d",&n);	printf("%lf\n",fact(n));	printf("%lf",factsum(n));	return 0;}double fact(int n){   	if(n == 1 || n == 0)		return 1;	else		return (double)n * fact(n-1);}double factsum(int n){   	int i;	double sum = 0;	for(i = 1;i <= n;i++)	{   		sum += fact(i);	}	return sum;}
   

3、递归指数函数

#include
   
    double calc_pow(double,int);int main(){   	int x,n;	scanf("%d,%d",&x,&n);	printf("x的n次方值为：%lf",calc_pow(x,n));}double calc_pow(double x,int n){   	if(n == 0)		return 1;	return x * calc_pow(x,n-1);}
   

4、交集密集函数

#include
   
    #include
    
     double fn(double x,int n);int main(){   	int x,n;	scanf("%d,%d",&x,&n);	printf("交错幂级函数值为:%lf",fn(x,n));	return 0;}double fn(double x,int n){   	if(n == 1)		return x;	else		return pow(-1,n-1) * pow(x,n) + fn(x,n-1);}
    
   

5、计算Ack函数

#include
   
    int Ack(int m,int n);int main(){   	int m,n;	scanf("%d,%d",&m,&n);	printf("%d",Ack(m,n));}int Ack(int m,int n){   	if(m > 0 && n > 0)		return Ack(m-1,Ack(m,n-1));	else if(n == 0 && m > 0)		return Ack(m-1,1);	else if(m == 0)		return n + 1;}
   

6、斐波拉契数列

#include
   
    int f(int n);int main(){   	int n;	scanf("%d",&n);	printf("%d",f(n));}int f(int n){   	if(n == 0)		return 0;	if(n == 1 || n == 2)		return 1;	else 		return f(n-1)+f(n-2);}
   

7、十进制转二进制

#include<stdio.h>

void dectobin(int n);int main(){   	int n;	scanf("%d",&n);	dectobin(n);    return 0;}void dectobin(int n){   	if(n == 0)		printf("0");	else if(n == 1)		printf("1");	else	{   		dectobin(n / 2);		printf("%d",n % 2);	}}

8、按顺序输出整数

#include
   
    void printdigits(int n);int main(){   	int n;	scanf("%d",&n);	printdigits(n);	return 0;}void printdigits(int n){   	if(n<10)	{   		printf("%d\n",n);	}	else	{   		printdigits(n / 10);		printf("%d\n",n % 10);	}}