PAT-1133 Splitting A Linked List（链表分解）

Cutting an integer means to cut a K digits long integer Z into two integers of (K/2) digits long integers A and B. For example, after cutting Z = 167334, we have A = 167 and B = 334. It is interesting to see that Z can be devided by the product of A and B, as 167334 / (167 x 334) = 3. Given an integer Z, you are supposed to test if it is such an integer.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<= 20). Then N lines follow, each gives an integer Z (10<=Z<=231). It is guaranteed that the number of digits of Z is an even number.

Output Specification:
For each case, print a single line "Yes" if it is such a number, or "No" if not.

Sample Input:
3
167334
2333

12345678

Sample Output:
Yes
No

No

题目大意：这道题形式与之前的乙级题反转链表很相似，给出一条链表以及一个整数k，要求将链表节点分为三部分输出：值小于0，值大于等于0小于等于k，值大于k，并且每部分节点的相对顺序不改变。

主要思路：先定义代表节点的结构体，包含地址（add），值（val），下个地址（next），获取输入并将每个节点存在其地址作为索引的数组中，然后从首节点开始遍历整个链表，如果节点值<0，则直接输出；如果<=k,则存入容器vec1中；如果>k，则存入容器vec2中，然后再依次遍历输出vec1和vec2中的节点。输出的时候注意，在第一次输出的时候只需要输出当前地址和节点值，其余时候需要输出两次地址（前一次作为上一个节点的next）和一次值。

#include 
#include 
using namespace std;
typedef struct{
    int add;
    int val;
    int next;
}Node;
Node node[100000]; 
vector vec1, vec2; 

int main(void) {
    int first, n, k, i;
	Node x;
	
    scanf("%d%d%d", &first, &n, &k);
    for (i = 0; i < n; i++) {
        scanf("%d%d%d", &x.add, &x.val, &x.next);    
        node[x.add] = x;			//节点放入其地址作为索引的数组中
    }

    //输出小于 0 的部分
	bool is_first = true;
    for (i = first; i != -1; i = node[i].next) {
        if (node[i].val < 0) {
            if (is_first) {
                printf("%05d %d ", node[i].add, node[i].val);
                is_first = false;
            }
            else
                printf("%05d\n%05d %d ", node[i].add, node[i].add, node[i].val);
        }
        else if (node[i].val <= k) 
            vec1.push_back(node[i]);
        else 
            vec2.push_back(node[i]);
    }
    
    //输出 [0, k] 的部分
    for (i = 0; i < vec1.size(); i++) {
        if (is_first) {
            printf("05d %d ", vec1[i].add, vec1[i].val);
            is_first = false;
        }
        else
            printf("%05d\n%05d %d ", vec1[i].add, vec1[i].add, vec1[i].val);
    }
       
    //输出大于 k 的部分
    for (i = 0; i < vec2.size(); i++) {
        if (is_first) {
            printf("%05d %d ", vec2[i].add, vec2[i].val);
            is_first = false;
        }
        else
            printf("%05d\n%05d %d ", vec2[i].add, vec2[i].add, vec2[i].val);
    }
    printf("-1\n");
    
    return 0;
}