课堂作业二 PAT1025 反转链表

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终于~奔溃了无数次后，看到这个结果 ，感动得不要不要的::>_<::

题目在这里

题目简述：该题可大致分为 输入链表 -> 链表节点反转 -> 两个步骤

输入链表：对于每个输入的值都要附带输入 存储该值的空间地址 和 下一个节点的地址，本来想用比较方便→_→但是只会不太会用，只能老老实实地用数组下标做地址名，建立结构体以存储附带数据。

输入即简化为对于地址为adr的节点，存储进a[adr]中。完成数组输入后，以第一个地址为起点，a[adr].afterstr 为线索，一直找到某个节点使其a[adr].afterstr==-1为止，循环开辟链表空间并输入。

链表节点反转：head->1->2->3->4->5->..->tail 我的方法是，三个指针 temp 、 q和p ，temp作为头指针，*p指向待反转的指针的前一位，假若k=4，从4开始 temp->4->1->2->3->5 , temp->4->3->1->2->5... temp->4->3->2->1->5，然后temp初始化，以1作为头指针的位置，重复上述循环。

链表输出：打代码时，输出一直忘了更新一下，输出应该为当前节点的下一个节点地址→_→ 并且输出的截止点不应该为p ->afterstr==-1 而是p ->next!= NULL （坑了一脸的血泪）

...然后就好了

注意：输入的节点可能不在链表上，因此要加一个计数器 count~

    #include<stdio.h>
    #include<stdlib.h>
    #include<iostream>
    using namespace std;

    struct MyStruct   //建立节点相关信息数据结构
   {
    int afterstr;
    int theadr;
    int num;
    MyStruct *next;
    };

    struct list     //创建链表
    {
     int afterstr;
     int num;
    }a[100001];

    int main()
    {
      int count = 0, times;
      int adr, n, k;
      int i, j;
      int theastr, afterstr, num;

      scanf("%d %d %d", &adr, &n, &k);     //输入头结点地址
      if (adr == -1)
      {
        printf("-1\n");
        return 0;
      }

      for (i = 0; i < n; i++)
      {
        scanf("%d %d %d", &theastr, &num, &afterstr);   //存入数组中
        a[theastr].afterstr = afterstr;
        a[theastr].num = num;
      }

      MyStruct *head, *pnew, *tail, *p, *q, *temp;
      head = (MyStruct*)malloc(sizeof(MyStruct));

      head->next = NULL;
      tail = head;

      for (theastr = adr; theastr != -1; theastr = a[theastr].afterstr)   //存入连表中
      {
        pnew = (MyStruct*)malloc(sizeof(MyStruct));       //开辟新节点
        pnew->num = a[theastr].num;
        pnew->afterstr = a[theastr].afterstr;
        pnew->theadr = theastr;
        tail->next = pnew;
        tail = pnew;
        count += 1;
      }

      times = count / k;

      p = head->next;
      temp = head;
      for (i = 0; i < times; i++)     //链表反转
      {
        for (j = 0; j < k - 1; j++)
        {

          q =p->next;
          p->next = q->next;
          q->next =temp->next;
          temp->next = q;

        }
        temp = p;     //重置头指针地址
       if (p->next == NULL)break;
        p = p->next;
      }

      for (p = head->next; p ->next!= NULL; p = p->next)    //打印链表
      {
        printf("%05d %d %05d\n", p->theadr,p->num,p->next->theadr);

      }
      printf("%05d %d -1\n", p->theadr, p->num);
      return 0;
    }

顺便，pta两题：

5-1

    #include<stdio.h>
    class Date
    {
    public:
        int year;
        int month;
        int day;
    };

    int main()
    {
        int monthday[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
        Date data;
    	int count;
        while (scanf("%d%d%d",&data.year,&data.month,&data.day)!=EOF)
        {
	        count = 0;
	        if (data.year == 0 && data.month == 0 && data.day == 0)
		        break;

        	if (data.year % 4 == 0 )
	        {
		        if (data.year % 100 == 0)
		        {
			        if (data.year % 400 == 0)
			        {
				        monthday[2] = 29;
			        }
		        }
		        else
		        {
			        monthday[2] = 29;
		        }

        	}

	        for (int i = 0; i < data.month; i++)
	        {
	        	count += monthday[i];
	        }
        	count += data.day;
	        printf("%d\n", count);
        }
        return 0;
    }

5-2

    #include<stdio.h>
    #include<string>
    #include<algorithm>
    #include<iostream>
    using namespace std;
    #define MAX 1000000

    class dormistry {
    public:
        string name;
        int height;
        int weight;
    };
    int dors[MAX] = { 0 };
    dormistry dor[MAX];
    int dornum[MAX];

    int main()
    {
        int i,j=0,n;
        string names;
        int nums;
        int heights;
    	int weights;

        scanf("%d", &n);
        for (i = 0; i < n; i++)
        {
	        cin >> nums >> names >> heights >> weights;
	        getchar();
	        //scanf_s("%d %s %d %d", &nums,&names, &heights, &weights);
    		if (dors[nums] == 0)
        	{
	        	dors[nums] = 1;
		        dor[nums].name = names;
        		dor[nums].height = heights;
	        	dor[nums].weight = weights;
	        	dornum[j++] = nums;

        	}
	        else if (dors[nums] == 1)
	        {
    			if (heights > dor[nums].height)
        		{
	        		dor[nums].name = names;
		        	dor[nums].height = heights;
    				dor[nums].weight = weights;
        		}
	        }

        }

        sort(dornum, dornum + j );

        for (i = 0; i < j; i++)
        {
	        printf("%06d ",dornum[i]);
	        cout<<dor[dornum[i]].name<< " " << dor[dornum[i]].height<< " " << dor[dornum[i]].weight<<endl;
        }
        return 0;
    }