Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.

## Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤$10^5$) which is the total number of nodes, and a positive K (≤N) which is the size of a block. The address of a node is a 5-digit nonnegative integer, and NULL is represented by −1.

Then N lines follow, each describes a node in the format:

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

## Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

00100 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218

71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1

## 解答

#include<iostream>
#include<unordered_map>
#include<vector>

using namespace std;

typedef struct {
}node;

int main()
{
unordered_map<int, node> list;
scanf("%d %d %d", &head, &n, &k);
for (int i = 0; i < n; i++) {
scanf("%d %d %d", &add, &data, &next);
}
vector<node> li, ans;
while (p != -1) {
li.push_back(list[p]);
p = list[p].next;
}
int re = li.size() % k, size = li.size();
int from = size - 1;
if (re != 0) {
for (int j = 0; j < re; j++)
ans.push_back(li[size - re + j]);
from -= re;
}
from -= k;
for (; from >= -1; from -= k) {
for (int j = 1; j <= k; j++)
ans.push_back(li[from + j]);
}
for (int i = 0; i < size - 1; i++)