蓝桥杯软件组C语言如何备赛，求解惑！

Question

各位仁人志士有没有建议，这个寒假应该如何备赛，想要以赛促学，今年刚学C语言半年。

Answer 1

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• 我还给你找了一篇非常好的博客，你可以看看是否有帮助，链接：C语言-编写一个函数，用来判断是否为素数，每输入一个数都调用一次此函数进行判断。
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• 除此之外, 这篇博客: 练习 用C语言编写一个程序，解释高等数学中的映射、单射和一一映射，并表示为函数。中的 练习 用C语言编写一个程序，解释高等数学中的映射、单射和一一映射，并表示为函数。 部分也许能够解决你的问题, 你可以仔细阅读以下内容或跳转源博客中阅读:
  

  代码块：

  #include <stdio.h>
  #include <stdlib.h>
  
  #define OK 1
  #define ERROR 0
  #define TRUE 1
  #define FALSE 0
  
  typedef int Status;
  typedef int Elemtype;
  
  typedef struct Set{
  	Elemtype num;
  	Set *next;
  	int length;
  }Set, *NumberSet;
  
  void input(NumberSet set);
  void output(NumberSet set);
  Status Function(Elemtype &x, Elemtype &y);
  Status Relationship1(NumberSet x, NumberSet y);
  Status Relationship2(NumberSet x, NumberSet y);
  Status Mapping(NumberSet x, NumberSet y, Status (*fun)(NumberSet x, NumberSet y));
  Status Injection(NumberSet x, NumberSet y, Status (*fun)(NumberSet x, NumberSet y));
  Status Bijection(NumberSet x, NumberSet y);
  
  void input(NumberSet set){
  	printf("Enter Set Length: ");
  	int len, i;
  	scanf_s("%d", &len);
  	printf("Enter Set %d Numbers: ", len);
  	Set *p = (Set*)malloc(sizeof(Set));
  	set->length = 0;
  	for(p = set, i = 0; i < len; i++, p = p->next){
  		scanf_s("%d", &p->num);
  		p->next = (Set*)malloc(sizeof(Set));
  		set->length++;
  		if(i == len - 1)
  			break;
  	}
  	p->next = NULL;
  }//input
  void output(NumberSet set){
  	printf("Set output: ");
  	Set *p;
  	for(p = set; p != NULL; p = p->next)
  		printf("%d ", p->num);
  	printf("\n");
  	printf("Set Length: %d\n", set->length);
  }//output
  Status Function(Elemtype &x, Elemtype &y){
  	if(y == 2*x)
  		return TRUE;
  	else
  		return FALSE;
  }//Function
  Status Relationship1(NumberSet x, NumberSet y){
  	Set *p, *q;
  	int count;
  	for(p = y, count = 0; p != NULL; p = p->next){
  		for(q = x; q != NULL; q = q->next)
  			if(Function(q->num, p->num)){
  				count++;
  				break;
  			}
  	}
  	if(count == y->length)
  		return TRUE;
  	else
  		return FALSE;
  }//Relationship1
  Status Relationship2(NumberSet x, NumberSet y){
  	Set *p, *q;
  	int count;
  	for(p = x, count = 0; p != NULL; p = p->next){
  		for(q = y; q !=	NULL; q = q->next)
  			if(Function(p->num, q->num)){
  				count++;
  				break;
  			}
  	}
  	if(count == x->length)
  		return TRUE;
  	else
  		return FALSE;
  }//Relationship2
  Status Mapping(NumberSet x, NumberSet y, Status (*fun)(NumberSet x, NumberSet y)){
  	fun = Relationship1;
  	if(fun(x, y))
  		return TRUE;
  	else
  		return FALSE;
  }//Mapping
  Status Injection(NumberSet x, NumberSet y, Status (*fun)(NumberSet x, NumberSet y)){
  	fun = Relationship2;
  	if(fun(x, y))
  		return TRUE;
  	else
  		return FALSE;
  }//Injection
  Status Bijection(NumberSet x, NumberSet y){
  	if(Mapping(x, y, Relationship1)&&Injection(x, y, Relationship2))
  		return TRUE;
  	else
  		return FALSE;
  }//Bijection
  
  int main()
  {
  	system("color a");
  
  	Set *numberset1 = (Set*)malloc(sizeof(Set));
  	Set *numberset2 = (Set*)malloc(sizeof(Set));
  	input(numberset1);
  	input(numberset2);
  	output(numberset1);
  	output(numberset2);
  	if(Mapping(numberset1, numberset2, Relationship1))
  		printf("Set1 to Set2 is Mapping\n");
  	if(Injection(numberset1, numberset2, Relationship2))
  		printf("Set1 to Set2 is Injection\n");
  	if(Bijection(numberset1, numberset2))
  		printf("Set1 to Set2 is Bijection\n");
  	free(numberset1);
  	free(numberset2);
  
  	system("pause");
  	return 0;
  }
  

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