Block reversal (summer vacation daily question 7)

Given a single chain table $L$, We will every $K$ A node is regarded as a block （ If the linked list is insufficient at the end $K$ Nodes , Also as a block ）, Please write a program to L The links of all blocks in are reversed .

for example ： Given $L$ by $1\to 2\to 3\to 4\to 5\to 6\to 7\to 8$,$K$ by $3$, Then the output should be $7\to 8\to 4\to 5\to 6\to 1\to 2\to 3$.

Add
This question may contain nodes that are not in the single linked list , These nodes need not be considered .

Input format
The first $1$ Line gives $1$ The address of the node 、 The total number of nodes is a positive integer $N$、 And positive integers $K$, That is, the size of the block . The address of the node is $5$ Bit nonnegative integer （ May contain leading $0$）,NULL Address use −1 Express .

Next there is $N$ That's ok , The format of each line is ：

Address Data Next

among Address Is the node address ,Data Is the integer data saved in this node ,Next Is the address of the next node .

Output format
For each test case , Output the inverted linked list in sequence , Each node on it occupies a row , The format is the same as the input .

Data range
$1\le K\le N\le 10^5$

sample 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

sample output ：

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

---

#include<iostream>
#include<vector>

using namespace std;

const int N = 100010;

int n, k;
int e[N], ne[N];

int main(){
    
    
    int h = -1;
    scanf("%d%d%d", &h, &n, &k);
    
    int addr, data, neaddr;
    for(int i = 0; i < n; i++){
    
        scanf("%d%d%d", &addr, &data, &neaddr);
        e[addr] = data;
        ne[addr] = neaddr;
    }
    
    vector<int> v, res;
    for(int i = h; i != -1; i = ne[i])
        v.push_back(i);
        
    int t = v.size() / k;
    if(v.size() % k != 0) t++;
    
    for(int i = t - 1; i >= 0; i--)
        for(int j = i * k; j < (i+1) * k && j < v.size(); j++)
            res.push_back(v[j]);
            
    for(int i = 0; i < res.size(); i++){
    
        
        printf("%05d %d ", res[i], e[res[i]]);
        if(i == res.size() - 1) puts("-1");
        else printf("%05d\n", res[i + 1]);
    }
            
    return 0;
}