使用递归/迭代思想完成n层嵌套平方根的计算

「爱情、让人受尽委屈。」 2022-10-22 05:26 125阅读 0赞

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70][]

\[程序代码(普通递归思想)\]

#include<stdio.h>
    #include<math.h>
    /* 函数原型 */
    double f(double x, unsigned int n);
    /* */
    int main()
    {
        double x;
        int n;
        printf("Please input x and n:");
        scanf("%lf,%d", &x, &n);
        if(x >=0 && n >= 0)
        {
            printf( "Result=%.2lf\n", f(x, n));
        }
        else
        {
            /* 数据非法 */
            printf("data error!\n");
        }
    }
    /* 计算 n层嵌套平方根 */
    double f(double x, unsigned int n)
    {
        if(n == 0)
        {
            /* 题目规定, 0层嵌套平方根的值为0 */
            return 0;
        }
        else if(n == 1)
        {
            /* 递归的终止条件 */
            return sqrt(x);
        }
        else
        {
            /* 递归的一般条件 */
            return sqrt(x + f(x, n - 1));
        }
    }

\[运行结果\]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 1][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 2][] ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 3][]

--------------------

另外也可以使用迭代思想解决该问题.

\[程序代码(迭代思想)\]

#include<stdio.h>
    #include<math.h>
    /* 函数原型 */
    double f(double x, unsigned int n);
    /* */
    int main()
    {
        double x;
        int n;
        printf("Please input x and n:");
        scanf("%lf,%d", &x, &n);
        if(x >=0 && n >= 0)
        {
            printf( "Result=%.2lf\n", f(x, n));
        }
        else
        {
            /* 数据非法 */
            printf("data error!\n");
        }
    }
    /* 计算 n层嵌套平方根 */
    double f(double x, unsigned int n)
    {
        if(n == 0)
        {
            /* 题目规定0层嵌套平方根的值为0 */
            return 0;
        }
        /* */
        double s = x;
        unsigned int j = 2;
        while(j <= n)
        {
            /* s变量负责实现迭代 */
            s = x + sqrt(s);
            j ++;
        }
        /* 循环结束后, 最后要取其平方根做为返回值 */
        s = sqrt(s);
        return s;
    }

\[运行结果\]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 4][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 5][]

--------------------

最后还可以通过尾递归思路解决该问题.

\[程序代码(尾递归思想)\]

#include<stdio.h>
    #include<math.h>
    /* 函数原型 */
    double f(double x, unsigned int n, double s);
    /* */
    int main()
    {
        double x;
        int n;
        printf("Please input x and n:");
        scanf("%lf,%d", &x, &n);
        if(x >=0 && n >= 0)
        {
            printf( "Result=%.2lf\n", f(x, n, x));
        }
        else
        {
            /* 数据非法 */
            printf("data error!\n");
        }
    }
    /* 计算 n层嵌套平方根 */
    double f(double x, unsigned int n, double s)
    {
        if(n == 0)
        {
            /* 题目规定0层嵌套平方根的值为0 */
            return 0;
        }
        else if(n == 1)
        {
            /* 递归的终止条件 */
            return sqrt(s);
        }
        else
        {
            /* 递归的一般条件 */
            /* 很容易通过迭代思路转化为尾递归思路 */
            s = x + sqrt(s);
            return f(x, n - 1, s);
        }
    }

\[运行结果\]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 6][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MjA0ODQ2Mw_size_16_color_FFFFFF_t_70 7][]

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