以多种不同的思想实现1+2+...+n的求和

[思路1(使用基本的递归思想完成计算)]

#include<stdio.h>
/* 函数原型 */
unsigned long Sum(unsigned int n);
/* */
int main()
{
    int n;
    printf("Please input n:");
    scanf("%d", &n);
    if(n >= 1)
    {
        printf("sum=%lu\n", Sum(n));
    }
    else
    {
        printf("data error!\n");
    }
    /* */
    return 0;
}
/* 递归求和函数 */
unsigned long Sum(unsigned int n)
{
    if(n == 1)
    {
        /* 递归的终止条件 */
        return 1;
    }
    else
    {
        /* 递归的一般条件 */
        return n + Sum(n - 1);
    }
}

[思路2(使用尾递归思想完成计算)]

#include<stdio.h>
/* 函数原型 */
unsigned long Sum(unsigned int n, unsigned long S);
/* */
int main()
{
    int n;
    printf("Please input n:");
    scanf("%d", &n);
    if(n >= 1)
    {
        printf("sum=%lu\n", Sum(n, 0));
    }
    else
    {
        printf("data error!\n");
    }
    /* */
    return 0;
}
/* 尾递归求和函数 */
unsigned long Sum(unsigned int n, unsigned long S)
{
    S = S + n;
    if(n == 1)
    {
        return S;
    }
    else
    {
        return Sum(n - 1, S);
    }
}

[思路3(使用迭代思想完成计算)]

#include<stdio.h>
/* 函数原型 */
unsigned long Sum(unsigned int n);
/* */
int main()
{
    int n;
    printf("Please input n:");
    scanf("%d", &n);
    if(n >= 1)
    {
        printf("sum=%lu\n", Sum(n));
    }
    else
    {
        printf("data error!\n");
    }
    /* */
    return 0;
}
/* 迭代求和函数 */
unsigned long Sum(unsigned int n)
{
    unsigned int j = 1;
    unsigned long S = 0;
    while(j <= n)
    {
        S = S + j;
        j ++;
    }
    return S;
}