Question

求用递归方法 写1到100的和

求用递归方法 写1到100的和

Answer 1

/*还没下班！喔，好冷啊！手都冻僵了!*/

#include<stdio.h>
int n=0;//递加后的结果，初始化为0。

int fun( int x)
{
n= x+n;
x--;
if(x<1)//这里不能用( x<=1)。
{
return n;
}
else
{ fun( x);};
}

main()
{
int i;
printf("请输入你要累加到的值\n");
scanf("%d",&i);
fun(i);
printf("累加后的值为:% d\n", n);

return 0;
}

Answer 2

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

int fun(int n)
{
if(n==0) return n;
else return n+fun(n-1);
}

void main()
{
int n; printf("***************SUM****************\n");
while(1){
printf("input n:\n");
scanf("%d",&n);printf("sum(%d) = %d\n",n,fun(n));
printf("continue,please input 1,or 0:\n");
scanf("%d",&n);
if(!n) break;
}
}

Answer 3

[cpp] view plain copy print?
#include<iostream>
using namespace std;
int sum(int max);
int main()
{
cout<<sum(100)<<endl;
return 0;
}
int sum(int max){
if(max>1){
return max+sum(max-1);
}else{
return 1;
}
}
[cpp] view plain copy print?
运行结果：5050

Answer 4

int func(int num)
{
int sum;
if (num > 1)
sum = num + func(num-1);
else
sum = 0;
return sum;
}

void main()
{
int sum = func(100);
printf("The sum of 1 to 100 is %d", sum);
return;
}