一道题目给了我疑惑

[矫情狗_____]

题目: (在线等...)
设计一个函数process,调用它时,每次实现不同的功能。(类似多态)
输入one和two两个数,第一个调用process时找出one和two的最大值,第二次找出最小值,第三次求和.

以下是我写的代码，process设计不出来...而且这个寻找函数首址的寻址方式有点头疼，求解释一下

#include <stdio.h>

void main()
{
	int Max(int one, int two);
	int Min(int one, int two);
	int Add(int one, int two);

	void process(int one, int two, int(*fan)());

	int one, two;
	scanf("%d%d", &one, &two);
	printf("Max = ");
	process(one, two, Max);
	printf("\nMin = ");
	process(one, two, Min);
	printf("\nAdd = ");
	process(one, two, Add);
	printf("\n");
}
int Max(int one, int two)
{
	int sum;
	if (one < two)
	{
		sum = two;
	}
	else
	{
		sum = one;
	}
	return sum;
}
int Min(int one, int two)
{
	int sum;
	if (one < two)
	{
		sum = one;
	}
	else
	{
		sum = two;
	}
	return sum;
}
int Add(int one, int two)
{
	int sum;
	sum = one + two;
        return sum;
}

void process(int one, int two, int(*fan)())
{
	int p1, p2, p3;

	p1 = (*fan)(one, two);


	p2 = (*fan)(one, two);


	p3 = (*fan)(one, two);
}

[bluewanderer]

int process(int one, int two)
{
	static int s = 0;

	int r = 0;

	switch (s)
	{
	case 0: r = one > two ? one : two; break;
	case 1: r = one < two ? one : two; break;
	case 2: r = one + two; break;
	}

	++s;

	return r;
}

为什么你给出的描述让人感觉是这种东西 +_+|||

[taoyh2002]

int process(int one, int two)
{
static int nCount = 0;
++nCount;

if (1 == nCount % 3)
{
return Max(one, two);
}
else if (2 == nCount % 3)
{
return Min(one, two);
}
else
{
return Add(one, two);
}
}

[矫情狗_____]

[Quote=引用 1 楼 bluewanderer 的回复:]
C/C++ code
int process(int one, int two)
{
static int s = 0;

int r = 0;

switch (s)
{
case 0: r = one > two ? one : two; break;
case 1: r = one < two ? one : two; br……
[/Quote]
我不确定是不是这个，题目就这么给我的...要求我做，但是写不明白。可是又没有答案，跟猫抓似的

[矫情狗_____]

有题目限制....

[hili210]

#include <stdio.h>

void main()
{
    int Max(int one, int two);
    int Min(int one, int two);
    int Add(int one, int two);

    int process(int one, int two, int(*fan)() );
    int one, two;


    scanf("%d%d", &one, &two);


    printf("Max = %d", process(one, two, Max) );
    printf("\nMin = %d", process(one, two, Min) );
    printf("\nAdd = %d", process(one, two, Add) );


    printf("\n");
}


int Max(int one, int two)
{
    int sum;
    if (one < two)
    {
        sum = two;
    }
    else
    {
        sum = one;
    }
    return sum;
}
int Min(int one, int two)
{
    int sum;
    if (one < two)
    {
        sum = one;
    }
    else
    {
        sum = two;
    }
    return sum;
}
int Add(int one, int two)
{
    int sum;
    sum = one + two;
        return sum;
}


int process(int one, int two, int(*fan)())
{
    return (*fan)(one, two);
}

[ncjobs2dick]

感谢hili210 ！