问题1_1

        函数 $fun$ 的功能是计算：
$$s=\sqrt{\ln (1)+\ln (2)+\ln (3)+\cdots +\ln (n)}$$
         $s$ 作为函数值返回。

代码1_1

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<math.h>double fun(int m){int i;double s=0.0;for(i=1; i<=m; i++)s = s+log(i);return sqrt(s);
}void main(void){system("CLS");printf("%f\n", fun(20));
}

结果1_1

问题1_2

        函数$fun$的功能是：求三个数的最小公倍数。

代码1_2

#include<stdio.h>int fun(int x, int y, int z){int j, t, n, m;j = 1;t = j%x;m = j%y;n = j%z;while(t!=0||m!=0||n!=0){j = j+1;t = j%x;m = j%y;n = j%z;}return j;
}void main(void){int x1, x2, x3, j;printf("Input x1 x2 x3:");scanf("%d %d %d", &x1, &x2, &x3);printf("x1 = %d, x2 = %d, x3 = %d\n", x1, x2, x3);j = fun(x1, x2, x3);printf("The minimal common mltiple is:%d\n", j);
}

结果1 _2

问题1_3

         函数$fun$的功能是：用递归算法求形参 $a$ 等待的平方根 。求平方根的迭代公式如下 。
        $x_1=\frac{1}{2}(x_0+\frac{a}{x_0})$
        迭代步骤如下：
         例如: $a$ 为 $2$ 时，平方根值为 $1.414214$ 。

代码1_3

#include<math.h>
#include<stdio.h>double fun(double a, double x0){double x1, y;x1 = (x0+a/x0)/2.0;if(fabs(x1-x0)>=0.00001)y = fun(a, x1);elsey = x1;return y;
}void main(void){double x;printf("Enter x:");scanf("%lf", &x);printf("The square root of %lf is %lf\n", x, fun(x, 1.0));
}

结果1_3