exec_outcome stringclasses 1 value | code_uid stringlengths 32 32 | file_name stringclasses 111 values | prob_desc_created_at stringlengths 10 10 | prob_desc_description stringlengths 63 3.8k | prob_desc_memory_limit stringclasses 18 values | source_code stringlengths 117 65.5k | lang_cluster stringclasses 1 value | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_time_limit stringclasses 27 values | prob_desc_sample_outputs stringlengths 2 796 | prob_desc_notes stringlengths 4 3k ⌀ | lang stringclasses 5 values | prob_desc_input_from stringclasses 3 values | tags listlengths 0 11 | src_uid stringlengths 32 32 | prob_desc_input_spec stringlengths 28 2.37k ⌀ | difficulty int64 -1 3.5k ⌀ | prob_desc_output_spec stringlengths 17 1.47k ⌀ | prob_desc_output_to stringclasses 3 values | hidden_unit_tests stringclasses 1 value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
PASSED | ceb8eea02796379c37bc0e7836278a41 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | //package Algorithm;
import java.awt.List;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.ListIterator;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;
import javax.xml.stream.events.StartDocument;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st = new StringTokenizer(br.readLine());
int T = Integer.parseInt(st.nextToken());
for (int test = 1; test <= T; test++) {
st = new StringTokenizer(br.readLine());
int N = Integer.parseInt(st.nextToken());
bw.write(N+"\n");
int[] seq = new int[N + 1];
for (int i = 1; i <= N; i++) {
seq[i] = i;
}
for(int i=1;i<=N;i++) {
bw.write(seq[i]+" ");
}
bw.write("\n");
for (int j = 2; j <= N; j++) {
int tmp = seq[j];
seq[j] = seq[j-1];
seq[j-1] = tmp;
for (int i = 1; i <= N; i++) {
bw.write(seq[i] + " ");
}
bw.write("\n");
}
}
bw.flush();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 98aad397caf711173d0b133e5e18d4d8 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class main
{
static long mod = (int)1e9+7;
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
// int t=1;
while(t-->0)
{
int n=sc.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++) {
arr[i]=i+1;
}
out.println(n);
for(int x:arr)
{
out.print(x+" ");
}
out.println();
for(int i=0;i<n-1;i++)
{
int temp = arr[i];
arr[i] = arr[n - 1];
arr[n - 1] = temp;
for(int x:arr)
{
out.print(x+" ");
}
out.println();
}
}
out.flush();
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
static class FenwickTree
{
//Binary Indexed Tree
//1 indexed
public int[] tree;
public int size;
public FenwickTree(int size)
{
this.size = size;
tree = new int[size+5];
}
public void add(int i, int v)
{
while(i <= size)
{
tree[i] += v;
i += i&-i;
}
}
public int find(int i)
{
int res = 0;
while(i >= 1)
{
res += tree[i];
i -= i&-i;
}
return res;
}
public int find(int l, int r)
{
return find(r)-find(l-1);
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
private static long mergeAndCount(int[] arr, int l,
int m, int r)
{
// Left subarray
int[] left = Arrays.copyOfRange(arr, l, m + 1);
// Right subarray
int[] right = Arrays.copyOfRange(arr, m + 1, r + 1);
int i = 0, j = 0, k = l;long swaps = 0;
while (i < left.length && j < right.length) {
if (left[i] < right[j])
arr[k++] = left[i++];
else {
arr[k++] = right[j++];
swaps += (m + 1) - (l + i);
}
}
while (i < left.length)
arr[k++] = left[i++];
while (j < right.length)
arr[k++] = right[j++];
return swaps;
}
// Merge sort function
private static long mergeSortAndCount(int[] arr, int l,
int r)
{
// Keeps track of the inversion count at a
// particular node of the recursion tree
long count = 0;
if (l < r) {
int m = (l + r) / 2;
// Total inversion count = left subarray count
// + right subarray count + merge count
// Left subarray count
count += mergeSortAndCount(arr, l, m);
// Right subarray count
count += mergeSortAndCount(arr, m + 1, r);
// Merge count
count += mergeAndCount(arr, l, m, r);
}
return count;
}
static void my_sort(long[] arr)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list);
for(int i=0;i<arr.length;i++)
{
arr[i] = list.get(i);
}
}
static void reverse_sorted(int[] arr)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list , Collections.reverseOrder());
for(int i=0;i<arr.length;i++)
{
arr[i] = list.get(i);
}
}
static int LowerBound(int a[], int x) { // x is the target value or key
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
static int UpperBound(ArrayList<Integer> list, int x) {// x is the key or target value
int l=-1,r=list.size();
while(l+1<r) {
int m=(l+r)>>>1;
if(list.get(m)<=x) l=m;
else r=m;
}
return l+1;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static class Queue_Pair implements Comparable<Queue_Pair> {
int first , second;
public Queue_Pair(int first, int second) {
this.first=first;
this.second=second;
}
public int compareTo(Queue_Pair o) {
return Integer.compare(o.first, first);
}
}
static void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
static void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
static boolean isPalindrome(String str)
{
// Pointers pointing to the beginning
// and the end of the string
int i = 0, j = str.length() - 1;
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str.charAt(i) != str.charAt(j))
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;
}
static boolean palindrome_array(char arr[], int n)
{
// Initialise flag to zero.
int flag = 0;
// Loop till array size n/2.
for (int i = 0; i <= n / 2 && n != 0; i++) {
// Check if first and last element are different
// Then set flag to 1.
if (arr[i] != arr[n - i - 1]) {
flag = 1;
break;
}
}
// If flag is set then print Not Palindrome
// else print Palindrome.
if (flag == 1)
return false;
else
return true;
}
static boolean allElementsEqual(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
static boolean allElementsDistinct(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]!=arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
public static void reverse_Long(long[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
long temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
static boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
static boolean isReverseSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] < a[i + 1]) {
return false;
}
}
return true;
}
static int[] rearrangeEvenAndOdd(int arr[], int n)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
int[] array = list.stream().mapToInt(i->i).toArray();
return array;
}
static long[] rearrangeEvenAndOddLong(long arr[], int n)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
long[] array = list.stream().mapToLong(i->i).toArray();
return array;
}
static boolean isPrime(long n)
{
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (long i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static long getSum(long n)
{
long sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n/10;
}
return sum;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcdLong(long a, long b)
{
if (b == 0)
return a;
return gcdLong(b, a % b);
}
static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static int countDigit(int n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 37e133b0421f58d378a33dd7f090229b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class MyClass {
public static void main(String args[])throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
int arr[]=new int[n];
System.out.println(n);
for(int i=0;i<n;i++) arr[i]=i+1;
for(int i=0;i<arr.length;i++) System.out.print(arr[i]+" ");
System.out.println();
for(int i=1;i<n;i++){
int temp=arr[i];
arr[i]=arr[i-1];
arr[i-1]=temp;
String s="";
for(int j=0;j<arr.length;j++) s+=arr[j]+" ";
System.out.println(s);
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 8cccd704aa964db5e9892d845a07296f | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class practice{
public static FastReader sc;
public static PrintWriter out;
static class FastReader{
BufferedReader br;
StringTokenizer str;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (str == null || !str.hasMoreElements()){
try{
str = new StringTokenizer(br.readLine());
}
catch (IOException lastMonthOfVacation){
lastMonthOfVacation.printStackTrace();
}
}
return str.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch (IOException lastMonthOfVacation){
lastMonthOfVacation.printStackTrace();
}
return str;
}
}
public static void main(String[] args) throws IOException{
out = new PrintWriter(new BufferedOutputStream(System.out));
sc = new FastReader();
int t = sc.nextInt();
for(int i = 0; i<t; i++){
int n = sc.nextInt();
int[] Array = new int[n+1];
out.println(n);
for(int j = 1; j<=n ; j++) {
Array[j] = j;
out.print(Array[j]+" ");
}
out.println();
for(int k = 1; k<n ; k++){
int temp = Array[k];
Array[k] = Array[k+1];
Array[k+1] = temp;
for(int p = 1; p<=n ; p++) out.print(Array[p]+" ");
out.println();
}
}
out.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 6e91c67cc06b9570c857da2f2f1ba85f | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;
public class weird_algrithm {
static BufferedWriter output = new BufferedWriter(
new OutputStreamWriter(System.out));
static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
static int mod = 1000000007;
static String toReturn = "";
static int steps = Integer.MAX_VALUE;
static int maxlen = 1000005;
/*MATHEMATICS FUNCTIONS START HERE
MATHS
MATHS
MATHS
MATHS*/
static long gcd(long a, long b) {
if(b == 0) return a;
else return gcd(b, a % b);
}
static long powerMod(long x, long y, int mod) {
if(y == 0) return 1;
long temp = powerMod(x, y / 2, mod);
temp = ((temp % mod) * (temp % mod)) % mod;
if(y % 2 == 0) return temp;
else return ((x % mod) * (temp % mod)) % mod;
}
static long modInverse(long n, int p) {
return powerMod(n, p - 2, p);
}
static long nCr(int n, int r, int mod, long [] fact, long [] ifact) {
return ((fact[n] % mod) * ((ifact[r] * ifact[n - r]) % mod)) % mod;
}
static boolean isPrime(long a) {
if(a == 1) return false;
else if(a == 2 || a == 3 || a== 5) return true;
else if(a % 2 == 0 || a % 3 == 0) return false;
for(int i = 5; i * i <= a; i = i + 6) {
if(a % i == 0 || a % (i + 2) == 0) return false;
}
return true;
}
static int [] seive(int a) {
int [] toReturn = new int [a + 1];
for(int i = 0; i < a; i++) toReturn[i] = 1;
toReturn[0] = 0;
toReturn[1] = 0;
toReturn[2] = 1;
for(int i = 2; i * i <= a; i++) {
if(toReturn[i] == 0) continue;
for(int j = 2 * i; j <= a; j += i) toReturn[j] = 0;
}
return toReturn;
}
static long [] fact(int a) {
long [] arr = new long[a + 1];
arr[0] = 1;
for(int i = 1; i < a + 1; i++) {
arr[i] = (arr[i - 1] * i) % mod;
}
return arr;
}
/*MATHS
MATHS
MATHS
MATHS
MATHEMATICS FUNCTIONS END HERE */
/*SWAP FUNCTION START HERE
SWAP
SWAP
SWAP
SWAP
*/
static void swap(int i, int j, long[] arr) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, int[] arr) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, String [] arr) {
String temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, char [] arr) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
/*SWAP
SWAP
SWAP
SWAP
SWAP FUNCTION END HERE*/
/*BINARY SEARCH METHODS START HERE
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
*/
static boolean BinaryCheck(long test, long [] arr, long health) {
for(int i = 0; i <= arr.length - 1; i++) {
if(i == arr.length - 1) health -= test;
else if(arr[i + 1] - arr[i] > test) {
health = health - test;
}else {
health = health - (arr[i + 1] - arr[i]);
}
if(health <= 0) return true;
}
return false;
}
static long binarySearchModified(long start1, long n, ArrayList<Long> arr, int a, long r) {
long start = start1, end = n, ans = -1;
while(start < end) {
long mid = (start + end) / 2;
//System.out.println(mid);
if(arr.get((int)mid) + arr.get(a) <= r && mid != start1) {
ans = mid;
start = mid + 1;
}else{
end = mid;
}
}
//System.out.println();
return ans;
}
static int binarySearch(int start, int end, ArrayList<Long> arr, long val) {
while(start < end) {
int mid = (start + end) / 2;
if(arr.get(mid) <= val) start = mid + 1;
else end = mid;
}
return start;
}
/*BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
BINARY SEARCH METHODS END HERE*/
/*RECURSIVE FUNCTION START HERE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
*/
static int recurse(int x, int y, int n, int steps1, Integer [][] dp) {
if(x > n || y > n) return 0;
if(dp[x][y] != null) {
return dp[x][y];
}
else if(x == n || y == n) {
return steps1;
}
return dp[x][y] = Math.max(recurse(x + y, y, n, steps1 + 1, dp), recurse(x, x + y, n, steps1 + 1, dp));
}
/*RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
RECURSIVE FUNCTION END HERE*/
/*GRAPH FUNCTIONS START HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
* */
static class edge{
int from, to;
long weight;
public edge(int x, int y, long weight2) {
this.from = x;
this.to = y;
this.weight = weight2;
}
}
static class sort implements Comparator<pair>{
@Override
public int compare(pair o1, pair o2) {
// TODO Auto-generated method stub
return (int)o1.a - (int)o2.a;
}
}
static void addEdge(ArrayList<ArrayList<edge>> graph, int from, int to, long weight) {
edge temp = new edge(from, to, weight);
edge temp1 = new edge(to, from, weight);
graph.get(from).add(temp);
//graph.get(to).add(temp1);
}
static int ans = 0;
static void topoSort(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, ArrayList<Integer> toReturn) {
if(visited[vertex]) return;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(!visited[graph.get(vertex).get(i)]) topoSort(graph, graph.get(vertex).get(i), visited, toReturn);
}
toReturn.add(vertex);
}
static boolean isCyclicDirected(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, boolean [] reStack) {
if(reStack[vertex]) return true;
if(visited[vertex]) return false;
reStack[vertex] = true;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(isCyclicDirected(graph, graph.get(vertex).get(i), visited, reStack)) return true;
}
reStack[vertex] = false;
return false;
}
static int e = 0;
static long mst(PriorityQueue<edge> pq, int nodes) {
long weight = 0;
int [] size = new int[nodes + 1];
Arrays.fill(size, 1);
while(!pq.isEmpty()) {
edge temp = pq.poll();
int x = parent(parent, temp.to);
int y = parent(parent, temp.from);
if(x != y) {
//System.out.println(temp.weight);
union(x, y, rank, parent, size);
weight += temp.weight;
e++;
}
}
return weight;
}
static void floyd(long [][] dist) { // to find min distance between two nodes
for(int k = 0; k < dist.length; k++) {
for(int i = 0; i < dist.length; i++) {
for(int j = 0; j < dist.length; j++) {
if(dist[i][j] > dist[i][k] + dist[k][j]) {
dist[i][j] = dist[i][k] + dist[k][j];
}
}
}
}
}
static void dijkstra(ArrayList<ArrayList<edge>> graph, long [] dist, int src) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MAX_VALUE / 2;
dist[src] = 0;
boolean visited[] = new boolean[dist.length];
PriorityQueue<pair> pq = new PriorityQueue<>(new sort());
pq.add(new pair(src, 0));
while(!pq.isEmpty()) {
pair temp = pq.poll();
int index = (int)temp.a;
for(int i = 0; i < graph.get(index).size(); i++) {
if(dist[graph.get(index).get(i).to] > dist[index] + graph.get(index).get(i).weight) {
dist[graph.get(index).get(i).to] = dist[index] + graph.get(index).get(i).weight;
pq.add(new pair(graph.get(index).get(i).to, graph.get(index).get(i).weight));
}
}
}
}
static int parent1 = -1;
static boolean ford(ArrayList<ArrayList<edge>> graph1, ArrayList<edge> graph, long [] dist, int src, int [] parent) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MIN_VALUE / 2;
dist[src] = 0;
boolean hasNeg = false;
for(int i = 0; i < dist.length - 1; i++) {
for(int j = 0; j < graph.size(); j++) {
int from = graph.get(j).from;
int to = graph.get(j).to;
long weight = graph.get(j).weight;
if(dist[to] < dist[from] + weight) {
dist[to] = dist[from] + weight;
parent[to] = from;
}
}
}
for(int i = 0; i < graph.size(); i++) {
int from = graph.get(i).from;
int to = graph.get(i).to;
long weight = graph.get(i).weight;
if(dist[to] < dist[from] + weight) {
parent1 = from;
hasNeg = true;
/*
* dfs(graph1, parent1, new boolean[dist.length], dist.length - 1);
* //System.out.println(ans); dfs(graph1, 0, new boolean[dist.length], parent1);
*/
//System.out.println(ans);
if(ans == 2) break;
else ans = 0;
}
}
return hasNeg;
}
/*GRAPH FUNCTIONS END HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
*/
/*disjoint Set START HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
static int [] rank;
static int [] parent;
static int parent(int [] parent, int x) {
if(parent[x] == x) return x;
else return parent[x] = parent(parent, parent[x]);
}
static boolean union(int x, int y, int [] rank, int [] parent, int [] setSize) {
if(parent(parent, x) == parent(parent, y)) {
return true;
}
if (rank[x] > rank[y]) {
parent[y] = x;
setSize[x] += setSize[y];
} else {
parent[x] = y;
setSize[y] += setSize[x];
if (rank[x] == rank[y]) rank[y]++;
}
return false;
}
/*disjoint Set END HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
/*INPUT START HERE
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT
*/
static int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(sc.readLine());
}
static long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(sc.readLine());
}
static long [] inputLongArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
long [] toReturn = new long[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Long.parseLong(s[i]);
}
return toReturn;
}
static int max = 0;
static int [] inputIntArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
//System.out.println(s.length);
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Integer.parseInt(s[i]);
}
return toReturn;
}
/*INPUT
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT END HERE
*/
static long [] preCompute(int level) {
long [] toReturn = new long[level];
toReturn[0] = 1;
toReturn[1] = 16;
for(int i = 2; i < level; i++) {
toReturn[i] = ((toReturn[i - 1] % mod) * (toReturn[i - 1] % mod)) % mod;
}
return toReturn;
}
static class pair{
long a;
long b;
long d;
public pair(long in, long y) {
this.a = in;
this.b = y;
this.d = 0;
}
}
static int [] nextGreaterBack(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int [] nextGreaterFront(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = s.length - 1; i >= 0; i--) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int lps(String s) {
int max = 0;
int [] lps = new int[s.length()];
lps[0] = 0;
int j = 0;
for(int i = 1; i < lps.length; i++) {
j = lps[i - 1];
while(j > 0 && s.charAt(i) != s.charAt(j)) j = lps[j - 1];
if(s.charAt(i) == s.charAt(j)) {
lps[i] = j + 1;
max = Math.max(max, lps[i]);
}
}
return max;
}
static int [][] vectors = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
static String dir = "DRUL";
static boolean check(int i, int j, boolean [][] visited) {
if(i >= visited.length || j >= visited[0].length) return false;
if(i < 0 || j < 0) return false;
return true;
}
static void selectionSort(long arr[], long [] arr1, ArrayList<ArrayList<Integer>> ans)
{
int n = arr.length;
for (int i = 0; i < n-1; i++)
{
int min_idx = i;
for (int j = i+1; j < n; j++)
if (arr[j] < arr[min_idx])
min_idx = j;
else if(arr[j] == arr[min_idx]) {
if(arr1[j] < arr1[min_idx]) min_idx = j;
}
if(i == min_idx) {
continue;
}
ArrayList<Integer> p = new ArrayList<Integer>();
p.add(min_idx + 1);
p.add(i + 1);
ans.add(new ArrayList<Integer>(p));
swap(i, min_idx, arr);
swap(i, min_idx, arr1);
}
}
static int saved = Integer.MAX_VALUE;
static String ans1 = "";
public static boolean isValid(int x, int y, String [] mat) {
if(x >= mat.length || x < 0) return false;
if(y >= mat[0].length() || y < 0) return false;
return true;
}
public static void recurse3(ArrayList<Character> arr, int index, String s, int max, ArrayList<String> toReturn) {
if(s.length() == max) {
toReturn.add(s);
return;
}
if(index == arr.size()) return;
recurse3(arr, index + 1, s + arr.get(index), max, toReturn);
recurse3(arr, index + 1, s, max, toReturn);
}
/*
if(arr[i] > q) return Math.max(f(i + 1, q - 1) + 1, f(i + 1, q);
else return f(i + 1, q) + 1
*/
static void dfsDP(ArrayList<ArrayList<Integer>> graph, int src, int [] dp1, int [] dp2, int parent) {
int sum1 = 0; int sum2 = 0;
for(int x : graph.get(src)) {
if(x == parent) continue;
dfsDP(graph, x, dp1, dp2, src);
sum1 += Math.min(dp1[x], dp2[x]);
sum2 += dp1[x];
}
dp1[src] = 1 + sum1;
dp2[src] = sum2;
System.out.println(src + " " + dp1[src] + " " + dp2[src]);
}
static int balanced = 0;
static void dfs(ArrayList<ArrayList<ArrayList<Long>>> graph, long src, int [] dist, long sum1, long sum2, long parent, ArrayList<Long> arr, int index) {
index = binarySearch(index, arr.size() - 1, arr, sum1);
if(index < arr.size() && arr.get(index) <= sum1) {
dist[(int)src] = index + 1;
}
else dist[(int)src] = index;
for(ArrayList<Long> x : graph.get((int)src)) {
if(x.get(0) == parent) continue;
if(arr.size() != 0) arr.add(arr.get(arr.size() - 1) + x.get(2));
else arr.add(x.get(2));
dfs(graph, x.get(0), dist, sum1 + x.get(1), sum2, src, arr, index);
arr.remove(arr.size() - 1);
}
}
static int compare(String s1, String s2) {
Queue<Character> q1 = new LinkedList<>();
Queue<Character> q2 = new LinkedList<Character>();
for(int i = 0; i < s1.length(); i++) {
q1.add(s1.charAt(i));
q2.add(s2.charAt(i));
}
int k = 0;
while(k < s1.length()) {
if(q1.equals(q2)) {
break;
}
q2.add(q2.poll());
k++;
}
return k;
}
static void solve() throws IOException {
int n = nextInt();
int [] arr = new int [n];
if(n == 1) {
output.write(1 + "\n");
}else {
output.write(n + "\n");
}
for(int i = 0; i < n; i++) {
arr[i] = i + 1;
output.write(i + 1 + " ");
}
output.write("\n");
int f = n;
int i = 0; int j = n - 1;
while(j >= 1) {
swap(j, j - 1, arr);
j--;
for(int k = 0; k < arr.length; k++) {
output.write(arr[k] + " ");
}
f -= 2;
output.write("\n");
}
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
int t = Integer.parseInt(sc.readLine()); for(int i = 0; i < t; i++)
solve();
output.flush();
}
}
class TreeNode {
int val;
TreeNode next;
public TreeNode(int x, TreeNode y) {
this.val = x;
this.next = y;
}
}
/*
1
10
6 10 7 9 11 99 45 20 88 31
*/
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | b5acd80fd717e1d8f82be228bf20d8c4 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class Solution{
public static void main(String args[]){
Scanner in = new Scanner(System.in);
long test = in.nextLong();
while(test-- > 0){
long n = in.nextLong();
ArrayList<Integer> list = new ArrayList<>();
int i, a = 1;
for(i = 0; i < n; i++){
list.add(i, i + 1);
}
ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
for(i = 0; i < n; i++){
ans.add(new ArrayList<>(list));
if(a >= n)
break;
else{
int temp = list.get(a);
list.set(a, list.get(0));
list.set(0, temp);
a++;
}
}
System.out.println(ans.size());
for(i = 0; i < n; i++){
System.out.println(Arrays.asList(ans.get(i)).toString().substring(1).replace("]", "").replace("[", "").replace(", ", " "));
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | e24d7bd7d0a1815c4154c4d5292f8896 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
import java.lang.*;
public class Codechef {
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
public static int cnt;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String args[]) {
int mod = 1000000007;
MyScanner scan = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
int test = scan.nextInt();
for(int t=0;t<test;t++){
int n = scan.nextInt();
int arr[] = new int[n];
out.println(n);
for(int i=0;i<n;i++){
out.print((i+1)+" ");
arr[i] = i+1;
}
out.println();
for(int i=0;i<n-1;i++){
int temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
for(int j=0;j<n;j++){
out.print(arr[j]+" ");
}
out.println();
}
}
out.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | fd742c86432a4a06da1ddeef3486a0d0 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.math.BigInteger;
public class Main{
static class Node {
long sum, pre;
Node(long a, long b) {
this.sum = a;
this.pre = b;
}
}
static class SegmentTree {
int l , r; // range responsible for
SegmentTree left , right;
long val;
SegmentTree(int l,int r,long a[]) {
this.l = l;
this.r = r;
// list= new ArrayList<>();
if(l == r) {
val = a[l];
return ;
}
int mid = l + (r-l)/2;
this.left = new SegmentTree(l ,mid , a);
this.right = new SegmentTree(mid + 1 , r,a);
this.val = Math.max(this.left.val , this.right.val);
}
public long query(int left ,int right) {
if(this.l > right || this.r < left) {
return 0l;
}
if(this.l >= left && this.r <= right) {
return this.val;
}
return this.left.query(left , right ) + this.right.query(left , right );
}
// public void pointUpdate(int index , long val) {
// if(this.l > index || this.r < index) return;
// if(this.l == this.r && this.l == index) {
// this.val = val;
// return ;
// }
// this.left.pointUpdate(index ,val);
// this.right.pointUpdate(index , val);
// this.val = join(this.left.val , this.right.val);
// }
// public void rangeUpdate(int left , int right,long val) {
// if(this.l > right || this.r < left) return ;
// if(this.l >= left && this.r <= right) {
// this.val += val;
// }
// if(this.l == this.r) return;
// this.left.rangeUpdate(left , right , val);
// this.right.rangeUpdate(left , right , val);
// }
// public long valueAtK(int k) {
// if(this.l > k || this.r < k) return 0;
// if(this.l == this.r && this.l == k) {
// return this.val;
// }
// return join(this.left.valueAtK(k) , this.right.valueAtK(k));
// }
public int join(int a ,int b) {
return a + b;
}
}
static class Hash {
long hash[] ,mod = (long)1e9 + 7 , powT[] , prime , inverse[];
Hash(char []s) {
prime = 131;
int n = s.length;
powT = new long[n];
hash = new long[n];
inverse = new long[n];
powT[0] = 1;
inverse[n-1] = pow(pow(prime , n-1 ,mod), mod-2 , mod);
for(int i = 1;i < n; i++ ) {
powT[i] = (powT[i-1]*prime)%mod;
}
for(int i = n-2; i>= 0;i-=1) {
inverse[i] = (inverse[i+1]*prime)%mod;
}
hash[0] = (s[0] - 'a' + 1);
for(int i = 1; i < n;i++ ) {
hash[i] = hash[i-1] + ((s[i]-'a' + 1)*powT[i])%mod;
hash[i] %= mod;
}
}
public long hashValue(int l , int r) {
if(l == 0) return hash[r]%mod;
long ans = hash[r] - hash[l-1] +mod;
ans %= mod;
// System.out.println(inverse[l] + " " + pow(powT[l], mod- 2 , mod));
ans *= inverse[l];
ans %= mod;
return ans;
}
}
static class ConvexHull {
Stack<Integer>stack;
Stack<Integer>stack1;
int n;
Point arr[];
ConvexHull(Point arr[]) {
n = arr.length;
this.arr = arr;
Arrays.sort(arr , (a , b)-> {
if(a.x == b.x) return (int)(b.y-a.y);
return (int)(a.x-b.x);
});
Point min = arr[0];
stack = new Stack<>();
stack1 = new Stack<>();
stack.push(0);
stack1.push(0);
Point ob = new Point(2,2);
for(int i =1;i < n;i++) {
if(stack.size() < 2) stack.push(i);
else {
while(stack.size() >= 2) {
int a = stack.pop() , b = stack.pop() ,c = i;
int dir = ob.cross(arr[b] , arr[a] , arr[c]);
if(dir < 0) {
stack.push(b);
stack.push(a);
stack.push(c);
break;
}
stack.push(b);
}
if(stack.size() < 2) {
stack.push(i);
}
}
}
for(int i =1;i < n;i++) {
if(stack1.size() < 2) stack1.push(i);
else {
while(stack1.size() >= 2) {
int a = stack1.pop() , b = stack1.pop() ,c = i;
int dir = ob.cross(arr[b] , arr[a] , arr[c]);
if(dir > 0) {
stack1.push(b);
stack1.push(a);
stack1.push(c);
break;
}
stack1.push(b);
}
if(stack1.size() < 2) {
stack1.push(i);
}
}
}
}
public List<Point> getPoints() {
boolean vis[] = new boolean[n];
List<Point> list = new ArrayList<>();
// for(int x : stack) {
// list.add(arr[x]);
// vis[x] = true;
// }
for(int x : stack1) {
// if(vis[x]) continue;
list.add(arr[x]);
}
return list;
}
}
public static class Suffix implements Comparable<Suffix> {
int index;
int rank;
int next;
public Suffix(int ind, int r, int nr) {
index = ind;
rank = r;
next = nr;
}
public int compareTo(Suffix s) {
if (rank != s.rank) return Integer.compare(rank, s.rank);
return Integer.compare(next, s.next);
}
}
static class Point {
long x , y;
Point(long x , long y) {
this.x = x;
this.y = y;
}
public String toString() {
return this.x + " " + this.y;
}
public Point sub(Point a,Point b) {
return new Point(a.x - b.x , a.y-b.y);
}
public int cross(Point a ,Point b , Point c) {
Point g = sub(b,a) , l = sub(c, b);
long ans = g.y*l.x - g.x*l.y;
if(ans == 0) return 0;
if(ans < 0) return -1;
return 1;
}
}
static class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio() { this(System.in, System.out); }
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio(String problemName) throws IOException {
super(new FileWriter(problemName + ".out"));
r = new BufferedReader(new FileReader(problemName + ".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) { }
return null;
}
public int nextInt() { return Integer.parseInt(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
public long nextLong() { return Long.parseLong(next()); }
}
static Kattio sc = new Kattio();
static long mod = 998244353l;
// static Kattio out = new Kattio();
//Heapify function to maintain heap property.
public static void swap(int i,int j,int arr[]) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static long max(long ...a) {
return maxArray(a);
}
public static void swap(int i,int j,long arr[]) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static void swap(int i,int j,char arr[]) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static String endl = "\n" , gap = " ";
static int size[];
static int parent[];
static HashMap<Integer , Long> value;
// static HashMap<String , Boolean > dp;
// static HashMap<Integer , List<int[]>> graph;
static boolean vis[];
static int answer;
static HashSet<String> guess;
static long primePow[];
// static long dp[];
static int N;
static int dis[];
static int height[];
static long p[];
// static long fac[];
// static long inv[];
static HashMap<Integer ,List<Integer>> graph;
public static long rd() {
return (long)((Math.random()*10) + 1);
}
public static void main(String[] args)throws IOException {
long t = ri();
// List<Integer> list = new ArrayList<>();
// int MAX = (int)4e4;
// for(int i =1;i<=MAX;i++) {
// if(isPalindrome(i + "")) list.add(i);
// }
// // System.out.println(list);
// long dp[] = new long[MAX +1];
// dp[0] = 1;
// long mod = (long)(1e9+7);
// for(int x : list) {
// for(int i =1;i<=MAX;i++) {
// if(i >= x) {
// dp[i] += dp[i-x];
// dp[i] %= mod;
// }
// }
// }
// int MAK = (int)1e6 + 10;
// boolean seive[] = new boolean[MAK];
// Arrays.fill(seive , true);
// seive[1] = false;
// seive[0] = false;
// for (int i = 1; i < MAK; i++) {
// if(seive[i]) {
// for (int j = i+i; j < MAK; j+=i) {
// seive[j] = false;
// }
// }
// }
// TreeSet<Long>primeSet = new TreeSet<>();
// for(long i = 2;i <= (long)1e6;i++) {
// if(seive[(int)i])primeSet.add(i);
// }
// List<Integer> list = new ArrayList<>();
// for (int i = 1; i < seive.length; i++) {
// if(seive[i]) list.add(i);
// }
int test_case = 1;
while(t-->0) {
// sc.print("Case #"+(test_case++)+": ");
solve();
}
sc.close();
}
static int endTime[];
static int time;
public static int dis(int a[] , int b[]) {
return (a[0]-b[0])*(a[0]-b[0]) + (a[1]-b[1])*(a[1]-b[1]);
}
static Long dp[][][];
static long fac[], inv[];
public static long ncr(int a , int b) {
if(a == b) return 1l;
return (((fac[a]*inv[b])%mod)*inv[a-b])%mod;
}
public static void solve() throws IOException {
List<int[]> l = new ArrayList<>();
int n = ri();
int a[] = new int[n];
for(int i =1 ;i <= n;i++) a[i-1] = i;
l.add(a);
for(int i = n-2;i >= 0;i--) {
int size = l.size()-1;
int b[] = l.get(size).clone();
swap(n-1 , i , b);
l.add(b);
}
sc.println(l.size());
for(int g[] : l) {
for(int x : g)sc.print(x + " ");
sc.println();
}
}
public static String slope(Point a , Point b) {
if((a.x-b.x) == 0) return "inf";
long n = a.y- b.y;
if(n == 0) return "0";
long m = a.x-b.x;
boolean neg = (n*m < 0?true:false);
n = Math.abs(n);
m = Math.abs(m);
long g = gcd(Math.abs(n),Math.abs(m));
n /= g;
m /= g;
String ans = n+"/"+m;
if(neg) ans = "-" + ans;
return ans;
}
public static int lis(int A[], int size) {
int[] tailTable = new int[size];
int len;
tailTable[0] = A[0];
len = 1;
for (int i = 1; i < size; i++) {
if (A[i] < tailTable[0]) tailTable[0] = A[i];
else if (A[i] > tailTable[len - 1]) tailTable[len++] = A[i];
else tailTable[CeilIndex(tailTable, -1, len - 1, A[i])] = A[i];
}
return len;
}
public static int CeilIndex(int A[], int l, int r, int key) {
while (r - l > 1) {
int m = l + (r - l) / 2;
if (A[m] >= key) r = m;
else l = m;
}
return r;
}
public static long unset(long n, long k) {
long ans = (n & (~(1l << (k - 1l))));
return ans;
}
public static long setKthBit(long n, long k) {
return ((1l << k) | n);
}
// public static void makeGraph(int s , int n,int m) {
// graph = new HashMap<>();
// for(int i = s;i<=n;i++) {
// graph.put(i , new ArrayList<>());
// }
// for(int i =0;i < m;i++) {
// int u = ri() , v = ri();
// graph.get(u).add(v);
// graph.get(v).add(u);
// }
// }
public static int lcs(char a[] , char b[]) {
int n = a.length , m = b.length;
int dp[][] = new int[n + 1][m + 1];
for(int i =1;i <= n;i++) {
for(int j =1;j <= m;j++) {
if(a[i-1] == b[j-1]) dp[i][j] = 1 + dp[i-1][j-1];
else dp[i][j] = Math.max(dp[i-1][j] , dp[i][j-1]);
}
}
return dp[n][m];
}
public static int find(int node) {
if(node == parent[node]) return node;
return parent[node] = find(parent[node]);
}
public static void merge(int a ,int b ) {
a = find(a);
b = find(b);
if(a == b) return;
if(size[a] >= size[b]) {
parent[b] = a;
size[a] += size[b];
// primePow[a] += primePow[b];
// primePow[b] = 0;
}
else {
parent[a] = b;
size[b] += size[a];
// primePow[b] += primePow[a];
// primePow[a] = 0;
}
}
public static void processPowerOfP(long arr[]) {
int n = arr.length;
arr[0] = 1;
long mod = (long)1e9 + 7;
for(int i =1;i<n;i++) {
arr[i] = arr[i-1]*51;
arr[i] %= mod;
}
}
public static long hashValue(char s[]) {
int n = s.length;
long powerOfP[] = new long[n];
processPowerOfP(powerOfP);
long ans =0;
long mod = (long)1e9 + 7;
for(int i =0;i<n;i++) {
ans += (s[i]-'a'+1)*powerOfP[i];
ans %= mod;
}
return ans;
}
public static void dfs(int r,int c,char arr[][]) {
int n = arr.length , m = arr[0].length;
arr[r][c] = '#';
for(int i =0;i<4;i++) {
int nr = r + colx[i] , nc = c + coly[i];
if(nr < 0 || nc < 0 || nc >= m || nr>=n) continue;
if(arr[nr][nc] == '#') continue;
dfs(nr,nc,arr);
}
}
public static double getSlope(int a , int b,int x,int y) {
if(a-x == 0) return Double.MAX_VALUE;
if(b-y == 0) return 0.0;
return ((double)b-(double)y)/((double)a-(double)x);
}
public static boolean collinearr(long a[] , long b[] , long c[]) {
return (b[1]-a[1])*(b[0]-c[0]) == (b[0]-a[0])*(b[1]-c[1]);
}
public static boolean isSquare(long sum) {
long root = (int)Math.sqrt(sum);
return root*root == sum;
}
public static int[] suffixArray(String s) {
int n = s.length();
Suffix[] su = new Suffix[n];
for (int i = 0; i < n; i++) {
su[i] = new Suffix(i, s.charAt(i) - '$', 0);
}
for (int i = 0; i < n; i++)
su[i].next = (i + 1 < n ? su[i + 1].rank : -1);
Arrays.sort(su);
int[] ind = new int[n];
for (int length = 4; length < 2 * n; length <<= 1) {
int rank = 0, prev = su[0].rank;
su[0].rank = rank;
ind[su[0].index] = 0;
for (int i = 1; i < n; i++) {
if (su[i].rank == prev && su[i].next == su[i - 1].next) {
prev = su[i].rank;
su[i].rank = rank;
}
else {
prev = su[i].rank;
su[i].rank = ++rank;
}
ind[su[i].index] = i;
}
for (int i = 0; i < n; i++) {
int nextP = su[i].index + length / 2;
su[i].next = nextP < n ?
su[ind[nextP]].rank : -1;
}
Arrays.sort(su);
}
int[] suf = new int[n];
for (int i = 0; i < n; i++)
suf[i] = su[i].index;
return suf;
}
public static boolean isPalindrome(String s) {
int i =0 , j = s.length() -1;
while(i <= j && s.charAt(i) == s.charAt(j)) {
i++;
j--;
}
return i>j;
}
// digit dp hint;
public static long callfun(String num , int N, int last ,int secondLast ,int bound ,long dp[][][][]) {
if(N == 1) {
if(last == -1 || secondLast == -1) return 0;
long answer = 0;
int max = (bound==1)?(num.charAt(num.length()-N)-'0') : 9;
for(int i = 0;i<=max;i++) {
if(last > secondLast && last > i) {
answer++;
}
if(last < secondLast && last < i) {
answer++;
}
}
return answer;
}
if(secondLast == -1 || last == -1 ){
long answer = 0l;
int max = (bound==1)?(num.charAt(num.length()-N)-'0') : 9;
for(int i =0;i<=max;i++) {
int nl , nsl , newbound = bound==0?0:i==max?1:0;
if(last == - 1&& secondLast == -1 && i == 0) {
nl = -1 ; nsl = -1;
}
else {
nl = i;nsl = last;
}
long temp = callfun(num , N-1 , nl , nsl ,newbound, dp);
answer += temp;
if(last != -1 && secondLast != -1 &&((last > secondLast && last > i)||(last < secondLast && last < i))) answer++;
}
return answer;
}
if(dp[N][last][secondLast][bound] != -1) return dp[N][last][secondLast][bound];
long answer = 0l;
int max = (bound==1)?(num.charAt(num.length()-N)-'0') : 9;
for(int i =0;i<=max;i++) {
int nl , nsl , newbound = bound==0?0:i==max?1:0;
if(last == - 1&& secondLast == -1 && i == 0) {
nl = -1 ; nsl = -1;
}
else {
nl = i;nsl = last;
}
long temp = callfun(num , N-1 , nl , nsl ,newbound,dp);
answer += temp;
if(last != -1 && secondLast != -1 &&((last > secondLast && last > i)||(last < secondLast && last < i))) answer++;
}
return dp[N][last][secondLast][bound] = answer;
}
public static Long callfun(int index ,int pair,int arr[],Long dp[][]) {
long mod = 998244353l;
if(index >= arr.length) return 1l;
if(dp[index][pair] != null) return dp[index][pair];
Long sum = 0l , ans = 0l;
if(arr[index]%2 == pair) {
return dp[index][pair] = callfun(index + 1,pair^1 , arr,dp)%mod + callfun(index + 1 ,pair , arr , dp)%mod;
}
else {
return dp[index][pair] = callfun(index + 1,pair , arr,dp)%mod;
}
// for(int i =index;i<arr.length;i++) {
// sum += arr[i];
// if(sum%2 == pair) {
// ans = ans + callfun(i + 1,pair^1,arr , dp)%mod;
// ans%=mod;
// }
// }
// return dp[index][pair] = ans;
}
public static boolean callfun(int index , int n,int neg , int pos , String s) {
if(neg < 0 || pos < 0) return false;
if(index >= n) return true;
if(s.charAt(0) == 'P') {
if(neg <= 0) return false;
return callfun(index + 1,n , neg-1 , pos , s.charAt(1) + "N");
}
else {
if(pos <= 0) return false;
return callfun(index + 1 , n , neg , pos-1 , s.charAt(1) + "P");
}
}
public static void getPerm(int n , char arr[] , String s ,List<String>list) {
if(n == 0) {
list.add(s);
return;
}
for(char ch : arr) {
getPerm(n-1 , arr , s+ ch,list);
}
}
public static int getLen(int i ,int j , char s[]) {
while(i >= 0 && j < s.length && s[i] == s[j]) {
i--;
j++;
}
i++;
j--;
if(i>j) return 0;
return j-i + 1;
}
public static int getMaxCount(String x) {
char s[] = x.toCharArray();
int max = 0;
int n = s.length;
for(int i =0;i<n;i++) {
max = Math.max(max,Math.max(getLen(i , i,s) , getLen(i ,i+1,s)));
}
return max;
}
public static double getDis(int arr[][] , int x, int y) {
double ans = 0.0;
for(int a[] : arr) {
int x1 = a[0] , y1 = a[1];
ans += Math.sqrt((x-x1)*(x-x1) + (y-y1)*(y-y1));
}
return ans;
}
public static boolean valid(String x ) {
if(x.length() == 0) return true;
if(x.length() == 1) return false;
char s[] = x.toCharArray();
if(x.length() == 2) {
if(s[0] == s[1]) {
return false;
}
return true;
}
int r = 0 , b = 0;
for(char ch : x.toCharArray()) {
if(ch == 'R') r++;
else b++;
}
return (r >0 && b >0);
}
public static void primeDivisor(HashMap<Long , Long >cnt , long num) {
for(long i = 2;i*i<=num;i++) {
while(num%i == 0) {
cnt.put(i ,(cnt.getOrDefault(i,0l) + 1));
num /= i;
}
}
if(num > 2) {
cnt.put(num ,(cnt.getOrDefault(num,0l) + 1));
}
}
public static boolean isSubsequene(char a[], char b[] ) {
int i =0 , j = 0;
while(i < a.length && j <b.length) {
if(a[i] == b[j]) {
j++;
}
i++;
}
return j >= b.length;
}
public static long fib(int n ,long M) {
if (n == 0) {
return 0;
} else if (n == 1) {
return 1;
} else {
long[][] mat = {{1, 1}, {1, 0}};
mat = pow(mat, n-1 , M);
return mat[0][0];
}
}
public static long[][] pow(long[][] mat, int n ,long M) {
if (n == 1) return mat;
else if (n % 2 == 0) return pow(mul(mat, mat , M), n/2 , M);
else return mul(pow(mul(mat, mat,M), n/2,M), mat , M);
}
static long[][] mul(long[][] p, long[][] q,long M) {
long a = (p[0][0]*q[0][0] + p[0][1]*q[1][0])%M;
long b = (p[0][0]*q[0][1] + p[0][1]*q[1][1])%M;
long c = (p[1][0]*q[0][0] + p[1][1]*q[1][0])%M;
long d = (p[1][0]*q[0][1] + p[1][1]*q[1][1])%M;
return new long[][] {{a, b}, {c, d}};
}
public static long[] kdane(long arr[]) {
int n = arr.length;
long dp[] = new long[n];
dp[0] = arr[0];
long ans = dp[0];
for(int i = 1;i<n;i++) {
dp[i] = Math.max(dp[i-1] + arr[i] , arr[i]);
ans = Math.max(ans , dp[i]);
}
return dp;
}
public static void update(int low , int high , int l , int r, int val , int treeIndex ,int tree[]) {
if(low > r || high < l || high < low) return;
if(l <= low && high <= r) {
System.out.println("At " +low + " and " + high + " ans ttreeIndex " + treeIndex);
tree[treeIndex] += val;
return;
}
int mid = low + (high - low)/2;
update(low , mid , l , r , val , treeIndex*2 + 1, tree);
update(mid + 1 , high , l , r , val , treeIndex*2 + 2 , tree);
}
// static int colx[] = {1 ,1, -1,-1 , 2,2,-2,-2};
// static int coly[] = {-2 ,2, 2,-2,1,-1,1,-1};
static int colx[] = {1 ,-1, 0,0 , 1,1,-1,-1};
static int coly[] = {0 ,0, 1,-1,1,-1,1,-1};
public static void reverse(char arr[]) {
int i =0 , j = arr.length-1;
while(i < j) {
swap(i , j , arr);
i++;
j--;
}
}
public static void reverse(long arr[]) {
int i =0 , j = arr.length-1;
while(i < j) {
swap(i , j , arr);
i++;
j--;
}
}
public static void reverse(int arr[]) {
int i =0 , j = arr.length-1;
while(i < j) {
swap(i , j , arr);
i++;
j--;
}
}
public static long inverse(long x , long mod) {
return pow(x , mod -2 , mod);
}
public static int maxArray(int arr[]) {
int ans = arr[0] , n = arr.length;
for(int i =1;i<n;i++) {
ans = Math.max(ans , arr[i]);
}
return ans;
}
public static long maxArray(long arr[]) {
long ans = arr[0];
int n = arr.length;
for(int i =1;i<n;i++) {
ans = Math.max(ans , arr[i]);
}
return ans;
}
public static int minArray(int arr[]) {
int ans = arr[0] , n = arr.length;
for(int i =0;i<n;i++ ) {
ans = Math.min(ans ,arr[i]);
}
return ans;
}
public static long minArray(long arr[]) {
long ans = arr[0];
int n = arr.length;
for(int i =0;i<n;i++ ) {
ans = Math.min(ans ,arr[i]);
}
return ans;
}
public static int sumArray(int arr[]) {
int ans = 0;
for(int x : arr) {
ans += x;
}
return ans;
}
public static long sumArray(long arr[]) {
long ans = 0;
for(long x : arr) {
ans += x;
}
return ans;
}
public static long rl() {
return sc.nextLong();
}
public static char[] rac() {
return sc.next().toCharArray();
}
public static String rs() {
return sc.next();
}
public static char rc() {
return sc.next().charAt(0);
}
public static int [] rai(int n) {
int ans[] = new int[n];
for(int i =0;i<n;i++) {
ans[i] = sc.nextInt();
}
return ans;
}
public static long [] ral(int n) {
long ans[] = new long[n];
for(int i =0;i<n;i++) {
ans[i] = sc.nextLong();
}
return ans;
}
public static int ri() {
return sc.nextInt();
}
public static int getValue(int num ) {
int ans = 0;
while(num > 0) {
ans++;
num = num&(num-1);
}
return ans;
}
public static boolean isValid(int x ,int y , int n,char arr[][],boolean visited[][][][]) {
return x>=0 && x<n && y>=0 && y <n && !(arr[x][y] == '#');
}
// public static Pair join(Pair a , Pair b) {
// Pair res = new Pair(Math.min(a.min , b.min) , Math.max(a.max , b.max) , a.count + b.count);
// return res;
// }
// segment tree query over range
// public static int query(int node,int l , int r,int a,int b ,Pair tree[] ) {
// if(tree[node].max < a || tree[node].min > b) return 0;
// if(l > r) return 0;
// if(tree[node].min >= a && tree[node].max <= b) {
// return tree[node].count;
// }
// int mid = l + (r-l)/2;
// int ans = query(node*2 ,l , mid ,a , b , tree) + query(node*2 +1,mid + 1, r , a , b, tree);
// return ans;
// }
// // segment tree update over range
// public static void update(int node, int i , int j ,int l , int r,long value, long arr[] ) {
// if(l >= i && j >= r) {
// arr[node] += value;
// return;
// }
// if(j < l|| r < i) return;
// int mid = l + (r-l)/2;
// update(node*2 ,i ,j ,l,mid,value, arr);
// update(node*2 +1,i ,j ,mid + 1,r, value , arr);
// }
public static long pow(long a , long b , long mod) {
if(b == 1) return a;
if(b == 0) return 1;
long ans = pow(a , b/2 , mod)%mod;
if(b%2 == 0) {
return (ans*ans)%mod;
}
else {
return ((ans*ans)%mod*a)%mod;
}
}
public static long pow(long a , long b ) {
if(b == 1) return a;
if(b == 0) return 1;
long ans = pow(a , b/2);
if(b%2 == 0) {
return (ans*ans);
}
else {
return ((ans*ans)*a);
}
}
public static boolean isVowel(char ch) {
if(ch == 'a' || ch == 'e'||ch == 'i' || ch == 'o' || ch == 'u') return true;
if((ch == 'A' || ch == 'E'||ch == 'I' || ch == 'O' || ch == 'U')) return true;
return false;
}
// public static int getFactor(int num) {
// if(num==1) return 1;
// int ans = 2;
// int k = num/2;
// for(int i = 2;i<=k;i++) {
// if(num%i==0) ans++;
// }
// return Math.abs(ans);
// }
public static int[] readarr()throws IOException {
int n = sc.nextInt();
int arr[] = new int[n];
for(int i =0;i<n;i++) {
arr[i] = sc.nextInt();
}
return arr;
}
public static boolean isPowerOfTwo (long x) {
return x!=0 && ((x&(x-1)) == 0);
}
public static boolean isPrime(long num) {
if(num==1) return false;
if(num<=3) return true;
if(num%2==0||num%3==0) return false;
for(long i =5;i*i<=num;i+=6) {
if(num%i==0 || num%(i+2) == 0) return false;
}
return true;
}
public static boolean isPrime(int num) {
// System.out.println("At pr " + num);
if(num==1) return false;
if(num<=3) return true;
if(num%2==0||num%3==0) return false;
for(int i =5;i*i<=num;i+=6) {
if(num%i==0 || num%(i+2) == 0) return false;
}
return true;
}
// public static boolean isPrime(long num) {
// if(num==1) return false;
// if(num<=3) return true;
// if(num%2==0||num%3==0) return false;
// for(int i =5;i*i<=num;i+=6) {
// if(num%i==0) return false;
// }
// return true;
// }
public static void allMultiple() {
// int MAX = 0 , n = nums.length;
// for(int x : nums) MAX = Math.max(MAX ,x);
// int cnt[] = new int[MAX + 1];
// int ans[] = new int[MAX + 1];
// for (int i = 0; i < n; ++i) cnt[nums[i]]++;
// for (int i = 1; i <= MAX; ++i) {
// for (int j = i; j <= MAX; j += i) ans[i] += cnt[j];
// }
}
public static long gcd(long a , long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public static int gcd(int a , int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public static int get_gcd(int a , int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
public static long get_gcd(long a , long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
// public static long fac(long num) {
// long ans = 1;
// int mod = (int)1e9+7;
// for(long i = 2;i<=num;i++) {
// ans = (ans*i)%mod;
// }
// return ans;
// }
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | b006d4f99f1f0b06c65060bebc53c3a2 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class aa
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
StringBuffer ans = new StringBuffer();
while(t-->0)
{
int n=sc.nextInt();
int a[] = new int[n];
for(int i = 0; i<n; i++)
{
a[i] = i+1;
}
ans.append(String.valueOf(n));
ans.append(System.getProperty("line.separator"));
for(int i = 0; i<n; i++)
{
for(int x:a)
{
ans.append(String.valueOf(x)+' ');
}
if(i!=n-1) ans.append(System.getProperty("line.separator"));;
int temp = a[i];
a[i] = a[n-1];
a[n-1] = temp;
}
if(t>0) ans.append(System.getProperty("line.separator"));;
}
System.out.println(ans);
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 4a625e5fbe7b62ce061331a8c0dcd30b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
// Problem Link:
// Runtime: O()
// Space: O()
public class B2 {
// Public Static Variables
// Private Static Methods Starts Here
private static class StringLT implements Comparator<String> {
@Override
public int compare(String o1, String o2) {
return o1.compareToIgnoreCase(o2);
}
}
private static class Pair<K, V>{
K key;
V value;
public Pair(K first, V second){
key = first;
value = second;
}
}
public static void main(String[] args) {
PrintWriter out = new PrintWriter(System.out);
Scanner sc = new Scanner(System.in);
// Code Starts Here
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
out.append(n + "\n");
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = i+1;
out.append(arr[i]+" ");
}
out.append("\n");
for(int i = 0; i < n-1; i++) {
int store = arr[i];
arr[i] = arr[i+1];
arr[i+1] = store;
for(int j = 0; j < n; j++) {
out.append(arr[j] + " ");
}
out.append("\n");
}
out.flush();
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 8c97bee8fd2c43e9c38a2dd2a501c0df | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
// Problem Link:
// Runtime: O()
// Space: O()
public class B {
// Public Static Variables
// Private Static Methods Starts Here
private static class StringLT implements Comparator<String> {
@Override
public int compare(String o1, String o2) {
return o1.compareToIgnoreCase(o2);
}
}
private static class Pair<K, V>{
K key;
V value;
public Pair(K first, V second){
key = first;
value = second;
}
}
public static void main(String[] args) throws IOException {
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
Scanner sc = new Scanner(System.in);
// Code Starts Here
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
out.append(n + "\n");
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = i+1;
out.append(arr[i]+" ");
}
out.append("\n");
for(int i = 0; i < n-1; i++) {
int store = arr[i];
arr[i] = arr[i+1];
arr[i+1] = store;
for(int j = 0; j < n; j++) {
out.append(arr[j] + " ");
}
out.append("\n");
}
out.flush();
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | c36b0c85c9eab246c5f0f5c92c210997 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static long mod = (int)1e9+7;
static PrintWriter out=new PrintWriter(new BufferedOutputStream(System.out));
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
// int t=1;
O : while(t-->0)
{
int n = sc.nextInt();
out.println(n);
for(int i=0;i<n;i++){
for(int j=1;j<=i;j++){
out.print((j+1)+" ");
}
out.print("1 ");
for(int j=i+1;j<n;j++){
out.print((j+1)+" ");
}
out.println();
}
}
out.flush();
}
static void printN()
{
System.out.println("NO");
}
static void printY()
{
System.out.println("YES");
}
static int findfrequencies(int a[],int n)
{
int count=0;
for(int i=0;i<a.length;i++)
{
if(a[i]==n)
{
count++;
}
}
return count;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
static int[] EvenOddArragement(int nums[])
{
int i1=0,i2=nums.length-1;
while(i1<i2){
while(nums[i1]%2==0 && i1<i2){
i1++;
}
while(nums[i2]%2!=0 && i2>i1){
i2--;
}
int temp=nums[i1];
nums[i1]=nums[i2];
nums[i2]=temp;
}
return nums;
}
static int gcd(int a, int b) {
while (b != 0) {
int t = a;
a = b;
b = t % b;
}
return a;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static int DigitSum(int n)
{
int r=0,sum=0;
while(n>=0)
{
r=n%10;
sum=sum+r;
n=n/10;
}
return sum;
}
static boolean checkPerfectSquare(int number)
{
double sqrt=Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static boolean isPowerOfTwo(int n)
{
if(n==0)
return false;
return (int)(Math.ceil((Math.log(n) / Math.log(2)))) == (int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
static boolean isPrime2(int n)
{
if (n <= 1)
{
return false;
}
if (n == 2)
{
return true;
}
if (n % 2 == 0)
{
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2)
{
if (n % i == 0)
{
return false;
}
}
return true;
}
static String minLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for(int i=0;i<n;i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[0];
}
static String maxLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for (int i = 0; i < n; i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[arr.length-1];
}
static class P implements Comparable<P> {
int i, j;
public P(int i, int j) {
this.i=i;
this.j=j;
}
public int compareTo(P o) {
return Integer.compare(i, o.i);
}
}
static class pair{
int i,j;
pair(int x,int y){
i=x;
j=y;
}
}
static int binary_search(int a[],int value)
{
int start=0;
int end=a.length-1;
int mid=start+(end-start)/2;
while(start<=end)
{
if(a[mid]==value)
{
return mid;
}
if(a[mid]>value)
{
end=mid-1;
}
else
{
start=mid+1;
}
mid=start+(end-start)/2;
}
return -1;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | cddb1f628f609cda3b4a855970c1cd72 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class Main {
public static String solve(int j, int n) {
String ans = "";
int k = 1;
for (int i = 0; i < n; i++) {
if (i == j) {
ans = ans + n + " ";
} else {
ans = ans + k + " ";
k++;
}
}
return ans + "\n";
}
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
Map<Integer, String> m1 = new HashMap<>();
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
if (m1.containsKey(n)) {
System.out.println(n);
System.out.println(m1.get(n));
continue;
}
List<Integer> l1 = new ArrayList<>();
System.out.println(n);
String ans = "";
for (int i = n - 1; i >= 0; i--) {
String p = solve(i, n);
ans = ans + p;
}
m1.put(n, ans);
System.out.println(ans);
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 61dbf71c8435a2056401fdeef209217c | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Scanner;
import java.util.stream.IntStream;
public class Solution {
private static void solve() {
for (int t = 0, T = i32(); t < T; t++) {
int n = i32();
int[] perm = IntStream.rangeClosed(1, n).toArray();
print(n);
print(perm);
for (int i = 1; i < perm.length; i++) {
swap(perm, i, i - 1);
print(perm);
}
}
}
static void swap(int[] array, int i, int j) {
int tmp = array[i];
array[i] = array[j];
array[j] = tmp;
}
static final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
static final PrintStream out = new PrintStream(new BufferedOutputStream(System.out));
public static void main(String[] args) { try (in; out) { solve(); } }
private static void print(String s) { out.println(s); }
private static void print(int i) { out.println(i); }
private static void print(int[] ints) { for (int i : ints) out.print(i + " "); out.println(); }
private static int i32() { return in.nextInt(); }
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 52ce5bf4b0755082e56462043bfe167d | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Scanner;
public class Solution {
private static void solve(Scanner in, PrintStream out) {
int t = in.nextInt();
while (t-- > 0) {
int[] perm = new int[in.nextInt()];
out.println(perm.length);
for (int i = 0; i < perm.length; i++) {
perm[i] = i + 1;
out.print(perm[i] + " ");
}
out.println();
for (int i = 1; i < perm.length; i++) {
swap(perm, i, i - 1);
for (int item : perm) {
out.print(item + " ");
}
out.println();
}
}
}
static void swap(int[] array, int i, int j) {
int tmp = array[i];
array[i] = array[j];
array[j] = tmp;
}
public static void main(String[] args) {
final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
final PrintStream out = new PrintStream(new BufferedOutputStream(System.out));
try (in; out) {
solve(in, out);
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | ca10e6e4dac9404af64b18316d16115b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static int mod = 1000000007;
static void read(int arr[], int start, int end, FastReader in) {
for (int i = start; i < end; i++) {
arr[i] = in.nextInt();
}
}
static int sumArr(int arr[]) {
int sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
}
return sum;
}
static final int MAX = 10000000;
static int prefix[] = new int[MAX + 1];
static void buildPrefix() {
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean prime[] = new boolean[MAX + 1];
Arrays.fill(prime, true);
for (int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= MAX; i += p) {
prime[i] = false;
}
}
}
// Build prefix array
prefix[0] = prefix[1] = 0;
for (int p = 2; p <= MAX; p++) {
prefix[p] = prefix[p - 1];
if (prime[p]) {
prefix[p]++;
}
}
}
static int query(int L, int R) {
return prefix[R] - prefix[L - 1];
}
static int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
static Long gcd(Long a, Long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
static int min(int arr[]) {
int min = Integer.MAX_VALUE;
for (int i = 0; i < arr.length; i++) {
min = Math.min(min, arr[i]);
}
return min;
}
public static void sort(int[] arr) {
ArrayList<Integer> ls = new ArrayList<>();
for (int x : arr) {
ls.add(x);
}
Collections.sort(ls);
for (int i = 0; i < arr.length; i++) {
arr[i] = ls.get(i);
}
}
public static void sortrev(int[] arr) {
ArrayList<Integer> ls = new ArrayList<>();
for (int x : arr) {
ls.add(x);
}
Collections.sort(ls,Collections.reverseOrder());
for (int i = 0; i < arr.length; i++) {
arr[i] = ls.get(i);
}
}
public static void sort(long[] arr) {
ArrayList<Long> ls = new ArrayList<>();
for (long x : arr) {
ls.add(x);
}
Collections.sort(ls);
for (int i = 0; i < arr.length; i++) {
arr[i] = ls.get(i);
}
}
static int max(int arr[]) {
int max = Integer.MIN_VALUE;
for (int i = 0; i < arr.length; i++) {
max = Math.max(max, arr[i]);
}
return max;
}
static long max(long arr[]) {
long max = Long.MIN_VALUE;
for (int i = 0; i < arr.length; i++) {
max = Math.max(max, arr[i]);
}
return max;
}
static int max(int a, int b) {
return Math.max(a, b);
}
static long max(long a, long b){
return (a>b)?a:b;
}
static int min(int a, int b) {
return Math.min(a, b);
}
static long min(long a, long b){
return Math.min(a, b);
}
public static boolean isPrime(long n) {
if (n < 2) {
return false;
}
if (n == 2 || n == 3) {
return true;
}
if (n % 2 == 0 || n % 3 == 0) {
return false;
}
long sqrtN = (long) Math.sqrt(n) + 1;
for (long i = 6L; i <= sqrtN; i += 6) {
if (n % (i - 1) == 0 || n % (i + 1) == 0) {
return false;
}
}
return true;
}
static class Pair {
int first;
int second;
Pair(int f, int s) {
first = f;
second = s;
}
@Override
public String toString() {
return "first: " + first + " second: " + second;
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
void read(int arr[]) {
for (int i = 0; i < arr.length; i++) {
arr[i] = nextInt();
}
}
void read(long arr[]) {
for (int i = 0; i < arr.length; i++) {
arr[i] = nextLong();
}
}
void read(char arr[][]){
for(int i=0;i<arr.length;i++){
String s=next();
for(int j=0;j<s.length();j++)
arr[i][j]=s.charAt(j);
}
}
void read(Pair arr[]) {
for (int i = 0; i < arr.length; i++) {
int x = nextInt();
int y = nextInt();
arr[i] = new Pair(x, y);
}
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void printarr(int arr[]) throws IOException {
println(Arrays.toString(arr));
}
public void printarr(long arr[]) throws IOException {
println(Arrays.toString(arr));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void close() throws IOException {
bw.close();
}
}
public static int count(String s, char c) {
int count = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == c) {
count++;
}
}
return count;
}
public static int stoi(String s) {
return Integer.parseInt(s);
}
public static String itos(int n) {
return Integer.toString(n);
}
public static int ctoi(char c) {
return (int) c - 48;
}
public static long countSetBits(long n) {
long count = 0;
while (n != 0) {
count += (n & 1); // check last bit
n >>= 1;
}
return count;
}
static long MOD = (long) (1e9 + 7);
static long powerLL(long x, long n) {
long result = 1;
while (n > 0) {
if (n % 2 == 1) {
result = result * x % MOD;
}
n = n / 2;
x = x * x % MOD;
}
return result;
}
static long countBits(long number) {
return (long) (Math.log(number)
/ Math.log(2) + 1);
}
static boolean isPowerOfTwo(long n) {
if (n == 0) {
return false;
}
while (n != 1) {
if (n % 2 != 0) {
return false;
}
n = n / 2;
}
return true;
}
public static int LCS(String s1, String s2){
int l1=s1.length();
int l2=s2.length();
int dp[][]=new int[l1+1][l2+1];
for(int i=0;i<=l1;i++){
for(int j=0;j<=l2;j++){
if(i==0 || j==0)
dp[i][j]=0;
else{
if(s1.charAt(i-1)==s2.charAt(j-1))
dp[i][j]=1+dp[i-1][j-1];
else
dp[i][j]=Math.max(dp[i-1][j], dp[i][j-1]);
}
}
}
// for(var v:dp)
// System.out.println(Arrays.toString(v));
return dp[l1][l2];
}
public static String LCSstring(String s1, String s2){
int l1=s1.length();
int l2=s2.length();
int dp[][]=new int[l1+1][l2+1];
for(int i=0;i<=l1;i++){
for(int j=0;j<=l2;j++){
if(i==0 || j==0)
dp[i][j]=0;
else{
if(s1.charAt(i-1)==s2.charAt(j-1))
dp[i][j]=1+dp[i-1][j-1];
else
dp[i][j]=Math.max(dp[i-1][j], dp[i][j-1]);
}
}
}
String ans="";
int i=l1,j=l2;
while(i>0 && j>0){
if(s1.charAt(i-1)==s2.charAt(j-1)){
ans=s1.charAt(i-1)+ans;
i--;
j--;
}
else if(dp[i-1][j]>dp[i][j-1]){
i--;
}
else
j--;
}
return ans;
}
public static int factorial(int n)
{
return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
public static long factorial(long n)
{
return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
public static int nCr(int n, int r,int a) {
return factorial(n) / (factorial(r) * factorial(n - r));
}
public static boolean[] sieve()
{
int n=10000000;
// Create a boolean array
// "prime[0..n]" and
// initialize all entries
// it as true. A value in
// prime[i] will finally be
// false if i is Not a
// prime, else true.
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static int maxSubArraySum(int a[])
{
int size = a.length;
int max_so_far = Integer.MIN_VALUE, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
public static int calc(int arr[],int n){
if(n<0 || n<arr[0])
return 0;
if(n>=arr[2])
return max(max(1+calc(arr,n-arr[2]),1+calc(arr,n-arr[1])),1+calc(arr,n-arr[0]));
else if(n>=arr[1])
return max(1+calc(arr,n-arr[1]),1+calc(arr,n-arr[0]));
else if(n>=arr[0])
return 1+calc(arr,n-arr[0]);
else
return 0;
}
public static int calc(String s1, String s2){
int arr1[]=new int[26];
int arr2[]=new int[26];
for(char c:s1.toCharArray())
arr1[c-'a']++;
for(char c:s2.toCharArray())
arr2[c-'a']++;
int sum=0;
for(int i=0;i<26;i++)
sum+=((arr2[i]-arr1[i])>0)?(arr2[i]-arr1[i]):0;
return sum;
}
public static boolean isSorted(int arr[]){
int sor[]=new int[arr.length];
for(int i=0;i<arr.length;i++)
sor[i]=arr[i];
sort(sor);
boolean check=true;
for(int i=0;i<arr.length;i++){
if(sor[i]!=arr[i]){
check=false;
break;
}
}
return check;
}
public static int find(int arr[], int n){
for(int i=0;i<arr.length;i++)
if(arr[i]==n)
return i;
return -1;
}
public static long calc(long n,long m, long mid,long k){
long ans=0;
for(int i=1;i<=n;i++)
ans+=Math.min(mid/i,m);
return ans;
}
public static boolean check(int arr[],int k){
int count=0;
for(int i=0;i<arr.length;i++)
if(arr[i]%(i+1) == 0)
count++;
return count==k;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static long lcm(long a, long b){
return (a/gcd(a,b))*b;
}
static int limit=1000000;
public static boolean same(String s){
boolean check=true;
for(int i=1;i<s.length();i++){
if(s.charAt(i)!=s.charAt(i-1)){
check=false;
break;
}
}
return check;
}
static int limit1=1000000000;
public static void printCount(int arr[],FastWriter out){
try{
for(int p:arr)
out.print(p+" ");
out.println("");
}
catch ( Exception E){
}
}
public static class item{
int len;
int cost;
item(int l, int c){
len=l;
cost=c;
}
@Override
public String toString() {
return "len: " + len + " cost: " + cost;
}
}
public static class comparator implements Comparator<item>{
public int compare(item i, item j){
if(i.len*i.cost != j.len*j.cost)
return (-(i.len*i.cost - j.len*j.cost));
else
return (-(i.len-j.len));
}
}
public static void cpy(int arr1[],int arr2[]){
for(int i=0;i<arr1.length;i++)
arr2[i]=arr1[i];
}
static int UpperBound(ArrayList<Integer> a, int x) {// x is the key or target value
int l=-1,r=a.size();
while(l+1<r) {
int m=(l+r)>>>1;
if(a.get(m)<=x) l=m;
else r=m;
}
return l+1;
}
static int LowerBound(ArrayList<Integer> a, int x) { // x is the target value or key
int l=-1,r=a.size();
while(l+1<r) {
int m=(l+r)>>>1;
if(a.get(m)>=x) r=m;
else l=m;
}
return r-1;
}
static int binarySearch(ArrayList<Integer> arr, int x)
{
int left = 0, right = arr.size() - 1;
while (left <= right)
{
int mid = left + (right - left) / 2;
// Check if x is present at mid
if (arr.get(mid) == x)
return mid;
// If x greater, ignore left half
if (arr.get(mid) < x)
left = mid + 1;
// If x is smaller, ignore right half
else
right = mid - 1;
}
// if we reach here, then element was
// not present
return -1;
}
static int upper_bound(ArrayList<Integer> arr, int key)
{
int upperBound = 0;
while (upperBound < arr.size()) {
// If current value is lesser than or equal to
// key
if (arr.get(upperBound) <= key)
upperBound++;
// This value is just greater than key
else{
// System.out.print("The upper bound of " + key + " is " + arr[upperBound] + " at index " + upperBound);
return upperBound;
}
}
return -1;
// System.out.print("The upper bound of " + key + " does not exist.");
}
public static void solve(FastReader in,FastWriter out){
try{
//-----------------------------------------
int n=in.nextInt();
out.println(n);
for(int i=0;i<n;i++){
int counter=2;
for(int j=0;j<n;j++){
if(j==i) out.print("1 ");
else out.print(counter++ +" ");
}
out.println("");
}
//-----------------------------------------
}
catch(Exception e){
e.printStackTrace();
}
}
public static void main(String[] args) {
try {
FastReader in = new FastReader();
FastWriter out = new FastWriter();
int tc=1;
tc=in.nextInt();
while(tc-- != 0)
solve(in,out);
out.close();
} catch (Exception e) {
e.printStackTrace();
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 9a1d7ac54ce2496444d0fab395963fbe | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class contestA {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
Arrays.fill(mk,false);
int t = scanner.nextInt();
while (t-->0){
int n = scanner.nextInt();
ac(n);
}
}
static boolean[] mk = new boolean[101];
static String[] ans = new String[101];
static void ac(int n){
if(n>100) return;
if(!mk[n]){
ans[n]=n+"\n";
for(int i=n;i>=1;--i){
for(int j=1;j<=n;++j){
if(j==i) ans[n]+=n+" ";
if(j<n) ans[n]+=j+" ";
}
ans[n]+= '\n';
}
mk[n] = true;
}
System.out.print(ans[n]);
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 63ea62efb7cf466e63877a45ab03816b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
BufferedWriter out=new BufferedWriter(new OutputStreamWriter(System.out));
int t;
t=sc.nextInt();
first:
while(t-->0)
{
long res,fill,place;
int n=sc.nextInt();
System.out.println(n);
for(int i=1;i<=n;i++)
{
StringBuilder str=new StringBuilder();
for(int j=2;j<=i;j++)
str.append(j+" ");
str.append(1+" ");
for(int j=i+1;j<=n;j++)
str.append(j+" ");
System.out.println(str);
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 2e01f9e2088112cfb1b32fa6f74fc046 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class _1716B{
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
StringBuilder sb = new StringBuilder();
for (int t = in.nextInt(); t>0; t--){
int n = in.nextInt();
sb.append(n).append("\n");
int[] num = new int[n+1];
for (int i = 1; i <= n; i++)
sb.append(num[i]=i).append(" ");
sb.append("\n");
for (int i = n-1; i >= 1; i--) {
num[i] = num[n];
num[n] = i;
for (int j = 1; j <= n; j++)
sb.append(num[j]).append(" ");
sb.append("\n");
}
}
System.out.println(sb);
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 946b3ad722d9759391e1c4276ab61721 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.BufferedWriter;
import java.math.BigInteger;
public class Solution{
public static void main(String[] args) throws IOException{
FastReader sc=new FastReader();
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int z=sc.nextInt();
for(int k=0;k<z;k++){
int a=sc.nextInt();
int res[]=new int[a];
for(int i=1;i<=a;i++){
res[i-1]=i;
}
System.out.println(a);
for(int i=0;i<a;i++){
for(int j=0;j<a;j++){
out.write(res[j]+" ");
}
out.write("\n");
out.flush();
if(i<a-1){
int temp=res[i];
res[i]=res[i+1];
res[i+1]=temp;
}
}
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 46fd19c7aae21c4c37b4ef0f52c63b2d | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.io.*;
import java.util.*;
public class MyClass {
public static void main(String args[])throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
int arr[]=new int[n];
System.out.println(n);
for(int i=0;i<n;i++) arr[i]=i+1;
for(int i=0;i<arr.length;i++) System.out.print(arr[i]+" ");
System.out.println();
for(int i=1;i<n;i++){
int temp=arr[i];
arr[i]=arr[i-1];
arr[i-1]=temp;
String s="";
for(int j=0;j<arr.length;j++) s+=arr[j]+" ";
System.out.println(s);
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | ff960d3f06475d19b30fb2381b9f7452 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | //codeforces
//package someAlgorithms;
import java.util.*;
import java.io.*;
import java.lang.*;
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
public static void main(String[] args) throws IOException{
String[] strNums = br.readLine().split(" ");
int t=Integer.parseInt(strNums[0]);
while(t-->0) {
//todo
String[] strNums1 = br.readLine().split(" ");
long n=Long.parseLong(strNums1[0]);
// long h=Long.parseLong(strNums1[1]);
// long m=Long.parseLong(strNums1[2]);
// String str = br.readLine();
Long[] arr = new Long[(int)n];
bw.write(n+"");//if apply \n then bw come in new lineand delete prev line without printing therefore use .flush to come to next line as it prints before coming to ext line!!
// StringTokenizer tk = new StringTokenizer(br.readLine().trim());
bw.newLine();
bw.flush();
for(int i=0;i<n;i++) {
arr[i]=(long)i+1;
bw.write(arr[i]+" ");
}
bw.newLine();
bw.flush();
int ind=0;
while(ind+1<n) {
long temp =arr[ind+1];
arr[ind+1]=arr[ind];
arr[ind]=temp;
ind++;
for(int i=0;i<n;i++) {
bw.write(arr[i]+" ");
}
bw.newLine();
bw.flush();
}
}
}
}
/*
3
3 2
3 1
1
4
2 4
3 1
4 1
*/
//class Pair{
// public long a=0;
// public long b=0;
// public Pair(long val,long id){
// this.a=val;
// this.b=id;
// }
//
//}
//class Pair{
// public long val;
// public long id;
// public Pair(long val,long id){
// this.val=val;
// this.id=id;
// }
//
//}
//class Node{
// public int val;
//// public Node left=null;
//// public Node right=null;
// ArrayList<Node> children = new ArrayList<>();
//
//}
//class Comp implements Comparator<Pair>{
// public int compare(Pair p1,Pair p2) {
// if(p1.val<p2.val) {
// return -1;
// }
// if(p1.val>p2.val){
// return 1;
// }
// else return 0; //MUST WRITE THIS RETURN 0 FOR EQUAL CASE SINCE GIVE RUNTIME ERROR IN SOME COMPLIERS
// }
//}
//if take gcd of whole array the take default gcd=0 and keep doing gcd of 2 elements of array!
//public static int gcd(int a, int b){
// if(b==0){
// return a;
// }
// return gcd(b,a%b);
//}
//ArrayList<Long>[] adjlist = new ArrayList[(int)n+1]; //array of arraylist
//for(int i=0;i<n;i++) {
// String[] strNums2 = br.readLine().split(" ");
// long a=Long.parseLong(strNums2[0]);
// long b=Long.parseLong(strNums2[1]);
//
// adjlist[(int)a].add(b);
// adjlist[(int)b].add(a);
//}
//int[][] vis = new int[(int)n+1][(int)n+1];
//OR can make list of list :-
//List<List<Integer>> adjlist = new ArrayList<>();
//for(int i=0;i<n;i++){
// adjlist.add(new ArrayList<>());
//}
//OR 1-D vis array
//int[] vis = new int[(int)n+1];
/*
Long[] arr = new Long[(int)n];
StringTokenizer tk = new StringTokenizer(br.readLine().trim());
for(int i=0;i<n;i++) {
arr[i]=Long.parseLong(tk.nextToken());
}
Long[][] arr = new Long[(int)n][(int)m];
for(int i=0;i<n;i++) {
String[] strNums2 = br.readLine().split(" ");
for(int j=0;j<m;j++) {
arr[i][j]=Long.parseLong(strNums2[j]);
}
}
4
4 4 3 2
4 4 4 3
Main m = new Main(); //no need since pair class main class ne niche banao
Pair p = m.new Pair(i,i+1);
li.add(p);
*/
//double num = 3.9634;
//double onum=num;
//num = (double)((int)(num*100))/100;
//double down = (num);
//float up =(float) down; //if take up as double then will get large value when add (3.96+0.01)!!
//if(down<onum) {
// up=(float)down+(float)0.01;
//// System.out.println(((float)3.96+(float)0.01));
//// System.out.println(3.96+0.01);//here both double, so output double , so here get large output other than 3.97!!
//}
////in c++ 3.96+0.01 is by default 3.97 but in java need to type cast to float to get this output!!
//System.out.println(down +" "+up);
/*
#include <iostream>
#include <string>
#include<vector>
#include<queue>
#include<utility>
#include<limits.h>
#include <unordered_set>
#include<algorithm>
using namespace std;
*/
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | f1f54f57557afdb6ae49f62cd72c3c82 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.math.BigInteger;
import java.util.*;
public class Main{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
int testcase=sc.nextInt();
for(int tc=1;tc<=testcase;tc++){
//System.out.print("Case #"+tc+": ");
int n = sc.nextInt();
System.out.println(n);
for (int i = 1; i <= n; i++) {
StringBuilder str = new StringBuilder();
for (int j = 2; j <= i; j++) {
str.append(j + " ");
}
str.append(1 + " ");
for (int j = i + 1; j <= n; j++)
str.append(j+" ");
System.out.println(str);
}
//sc.close();
}
}
static long minSum(int arr[], int n, int k)
{
// k must be smaller than n
if (n < k)
{
// System.out.println("Invalid");
return -1;
}
// Compute sum of first window of size k
long res = 0;
for (int i=0; i<k; i++)
res += arr[i];
// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window.
long curr_sum = res;
for (int i=k; i<n; i++)
{
curr_sum += arr[i] - arr[i-k];
res = Math.min(res, curr_sum);
}
return res;
}
static long gcd(long a, long b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
static int nextIndex(int a[], int x){
int n=a.length;
for(int i=x;i<n-1;i++){
if(a[i]>a[i+1]){
return i;
}
}
return n;
}
static void rev(int a[], int i, int j){
while(i<j){
int t=a[i];
a[i]=a[j];
a[j]=t;
i++;
j--;
}
}
static int sorted(int arr[], int n)
{
// Array has one or no element or the
// rest are already checked and approved.
if (n == 1 || n == 0)
return 1;
// Unsorted pair found (Equal values allowed)
if (arr[n - 1] < arr[n - 2])
return 0;
// Last pair was sorted
// Keep on checking
return sorted(arr, n - 1);
}
static void sieveOfEratosthenes(int n, Set<Integer> set)
{
// Create a boolean array "prime[0..n]" and
// initialize all entries it as true. A value in
// prime[i] will finally be false if i is Not a
// prime, else true.
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true) {
// Update all multiples of p greater than or
// equal to the square of it numbers which
// are multiple of p and are less than p^2
// are already been marked.
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (int i = 2; i <= n; i++) {
if (prime[i] == true)
set.add(i);
}
}
static boolean isPowerOfTwo(int n)
{
return (int)(Math.ceil((Math.log(n) / Math.log(2))))
== (int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
static int countBits(int number)
{
// log function in base 2
// take only integer part
return (int)(Math.log(number) /
Math.log(2) + 1);
}
public static void swap(int ans[], int i, int j) {
int temp=ans[i];
ans[i]=ans[j];
ans[j]=temp;
}
static int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 59d2df7333fdc4dc57bc557f3a141b2c | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class A_1 {
static FastReader in=new FastReader();
static final Random random=new Random();
static long mod=1000000007L;
static HashMap<String,Integer>map=new HashMap<>();
public static void main(String args[]) throws IOException
{
int T = in.nextInt();
StringBuilder res=new StringBuilder();
for(int f = 1; f<=T; f++)
{
int p = in.nextInt();
res.append(p + "\n");
for(int i=0; i<p; i++)
{
for(int j=0; j<p-i-1; j++)
{
res.append((j+1) + " ");
}
res.append(p + " ");
for(int j=p-i; j<p; j++)
{
res.append(j + " ");
}
res.append("\n");
}
}
print(res);
}
static int max(int a, int b)
{
if(a<b)
return b;
return a;
}
static void ruffleSort(int[] a) {
int n=a.length;
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static < E > void print(E res)
{
System.out.println(res);
}
static int gcd(int a,int b)
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static int abs(int a)
{
if(a<0)
return -1*a;
return a;
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
int [] readintarray(int n) {
int res [] = new int [n];
for(int i = 0; i<n; i++)res[i] = nextInt();
return res;
}
long [] readlongarray(int n) {
long res [] = new long [n];
for(int i = 0; i<n; i++)res[i] = nextLong();
return res;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | b49b68a98c579faae8964868d56ef61d | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main
{
public static void main(String[] args)throws Exception {
Scanner sc=new Scanner(System.in);
int x=sc.nextInt();
while(x-->0)
{
int a=sc.nextInt();
int arr[]=new int[a];
BufferedWriter output =new BufferedWriter(new OutputStreamWriter(System.out));
for(int i=0;i<a;i++)
{
arr[i]=i+1;
}
output.write(a+"\n");
output.write(""+arr[0]);
for(int i=1;i<a;i++)
{
output.write(" "+arr[i]);
}
output.write("\n");
for(int i=0;i<a-1;i++)
{
arr[i]=arr[i]^arr[i+1];
arr[i+1]=arr[i]^arr[i+1];
arr[i]=arr[i]^arr[i+1];
output.write(""+arr[0]);
for(int j=1;j<a;j++)
{
output.write(" "+arr[j]);
}
output.write("\n");
}
output.flush();
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 264f8206b85a4914933ef38b92ee55b6 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class hello {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
try {
br = new BufferedReader(new FileReader("input.txt"));
PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
System.setOut(out);
}
// Catch block to handle the exceptions
catch (Exception e) {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void getans(int s,int i,String ans[],String str,int count)
{
if(s==0)
{
if(ans[0]!=null&&count<ans[0].length())
{
ans[0] = new String(str);
}
else if(ans[0]==null)
{
ans[0] = new String(str);
}
return;
}
if(s<i) return;
for(int j=i;j<=9;j++)
{
getans(s-j,j+1,ans,str+Integer.toString(j),count+1);
// System.out.println(j);
}
}
public static void main (String[] args) throws java.lang.Exception {
try {
FastReader sc = new FastReader();
BufferedWriter out = new BufferedWriter(
new OutputStreamWriter(System.out));
BufferedWriter output = new BufferedWriter(
new OutputStreamWriter(System.out));
int t = sc.nextInt();
while(t-->0)
{
int n = sc.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++) arr[i]=i+1;
System.out.println(n);
for(int i=0;i<n;i++)
{
output.write(arr[i]+" ");
}
output.write("\n");
int k = 1;
while(k<n)
{
int temp = arr[k-1];
arr[k-1] = arr[k];
arr[k] = temp;
for(int val:arr)
{
output.write(val+" ");
}
output.write("\n");
k++;
}
output.flush();
}
} catch (Exception e) {
return;
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | baf8b4df6c7c7388d06780ce351666e4 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
int T=0;
Scanner in = new Scanner(System.in);
T = in.nextInt();
while(T > 0) {
int n=0;
n=in.nextInt();
int[] arr = new int[n];
for(int i=0; i<n; i++) {
arr[i] =i+1;
}
System.out.println(n);
for(int i=0; i<n; i++) {
System.out.print(arr[i]+" ");
}
System.out.println();
int j=1, temp=0;
while(j<n) {
temp = arr[0];
arr[0] = arr[j];
arr[j] = temp;
String s="";
s = Arrays.toString(arr);
s=s.replace("[", "");
s=s.replace("]", "");
s=s.replace(",", " ");
System.out.println(s);
// for(int i=0; i<n ;i++) {
// System.out.print(arr[i]+" ");
// }
// System.out.println();
j++;
}
T--;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 6b52d59e8d0c0c0ac9fc18b2e5f61e37 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.StringTokenizer;
public class codeforces {
static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) throws IOException {
StringTokenizer st = new StringTokenizer(r.readLine());
int t = Integer.parseInt(st.nextToken());
for (int i = 0; i < t; i++) { // for each test case
st = new StringTokenizer(r.readLine());
int n = Integer.parseInt(st.nextToken());
pw.println(n);
int[] arr = new int[n];
for (int j = 0; j < n; j++) {
arr[j] = j+1;
pw.print(j+1+" ");
}
pw.println();
for (int j = 0; j < n-1; j++) {
int first = arr[j];
int second = arr[j+1];
arr[j+1] = first;
arr[j] = second;
for(int num : arr) {
pw.print(num+" ");
}
pw.println();
}
/*
if (n % 2 == 1) {
int first = arr[n-2];
int second = arr[n-1];
arr[n-1] = first;
arr[n-2] = second;
for(int num : arr) {
pw.print(num+" ");
}
pw.println();
}
*/
}
pw.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 100bd0a8965ba8b15ed409aaf0612c9f | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void print(int arr[],BufferedWriter log) throws Exception {
for(int i=0;i<arr.length;i++) {
log.write(arr[i]+" ");
}
log.write("\n");
}
public static void main(String[] args) throws Exception {
// TODO Auto-generated method stub
Reader s=new Reader();
//FastReader s=new FastReader();
//Scanner s=new Scanner(System.in);
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
int t=s.nextInt();
for(int x=0;x<t;x++) {
int n=s.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++) {
arr[i]=i+1;
}
log.write(n+"\n");
print(arr,log);
for(int i=0;i<n-1;i++) {
arr[i]=arr[i+1];
arr[i+1]=1;
print(arr,log);
}
}
log.flush();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 33b3520102245be9b10aeb6be1892c95 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
public class Round12 {
public static void main(String[] args) {
FastReader fastReader = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = fastReader.nextInt();
while (t-- > 0) {
int n = fastReader.nextInt();
int a[] = new int[n];
for (int i = 0; i < n; i++) a[i] = (i + 1);
out.println(n);
for (int i = n - 1; i >= 0; i--) {
if (i != n - 1){
int temp = a[i];
a[i] = a[i + 1];
a[i + 1] = temp;
}
StringBuilder ans = new StringBuilder();
for (int j = 0; j < n; j++) {
ans.append(a[j]).append(" ");
}
out.println(ans);
}
}
out.close();
}
public static void nextPermutation(int[] nums) {
int i = nums.length - 2;
while (i >= 0 && nums[i + 1] <= nums[i]) {
i--;
}
if (i >= 0) {
int j = nums.length - 1;
while (nums[j] <= nums[i]) {
j--;
}
swap(nums, i, j);
}
reverse(nums, i + 1);
}
public static void reverse(int[] nums, int start) {
int i = start, j = nums.length - 1;
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}
public static void swap(int[] nums, int j, int i) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
static class Pair {
int len;
int x;
int y;
Pair(int len, int x, int y) {
this.len = len;
this.x = x;
this.y = y;
}
}
// constants
static final int IBIG = 1000000007;
static final int IMAX = 2147483647;
static final long LMAX = 92233720368547L;
static Random __r = new Random();
// math util
static int minof(int a, int b, int c) {
return min(a, min(b, c));
}
static int minof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
int min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static long minof(long a, long b, long c) {
return min(a, min(b, c));
}
static long minof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
long min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static int maxof(int a, int b, int c) {
return max(a, max(b, c));
}
static int maxof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
int max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static long maxof(long a, long b, long c) {
return max(a, max(b, c));
}
static long maxof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
long max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static int powi(int a, int b) {
if (a == 0) return 0;
int ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static long powl(long a, int b) {
if (a == 0) return 0;
long ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static int fli(double d) {
return (int) d;
}
static int cei(double d) {
return (int) ceil(d);
}
static long fll(double d) {
return (long) d;
}
static long cel(double d) {
return (long) ceil(d);
}
static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
static int[] exgcd(int a, int b) {
if (b == 0) return new int[]{1, 0};
int[] y = exgcd(b, a % b);
return new int[]{y[1], y[0] - y[1] * (a / b)};
}
static long[] exgcd(long a, long b) {
if (b == 0) return new long[]{1, 0};
long[] y = exgcd(b, a % b);
return new long[]{y[1], y[0] - y[1] * (a / b)};
}
static int randInt(int min, int max) {
return __r.nextInt(max - min + 1) + min;
}
static long mix(long x) {
x += 0x9e3779b97f4a7c15L;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L;
x = (x ^ (x >> 27)) * 0x94d049bb133111ebL;
return x ^ (x >> 31);
}
public static boolean[] findPrimes(int limit) {
assert limit >= 2;
final boolean[] nonPrimes = new boolean[limit];
nonPrimes[0] = true;
nonPrimes[1] = true;
int sqrt = (int) Math.sqrt(limit);
for (int i = 2; i <= sqrt; i++) {
if (nonPrimes[i]) continue;
for (int j = i; j < limit; j += i) {
if (!nonPrimes[j] && i != j) nonPrimes[j] = true;
}
}
return nonPrimes;
}
// array util
static void reverse(int[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
int swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(long[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
long swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(double[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
double swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
char swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void shuffle(int[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
int swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(long[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
long swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(double[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
double swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void rsort(int[] a) {
shuffle(a);
sort(a);
}
static void rsort(long[] a) {
shuffle(a);
sort(a);
}
static void rsort(double[] a) {
shuffle(a);
sort(a);
}
static int[] copy(int[] a) {
int[] ans = new int[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static long[] copy(long[] a) {
long[] ans = new long[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static double[] copy(double[] a) {
double[] ans = new double[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static char[] copy(char[] a) {
char[] ans = new char[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] ria(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = Integer.parseInt(next());
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] rla(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = Long.parseLong(next());
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 3ed1a526985bbd9d998783de12c18f6e | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.util.*;
public class Codeforces {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc= new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
StringBuilder sb= new StringBuilder("");
int n=sc.nextInt();
System.out.println(n);
for(int i=0;i<n;i++)
{
sb.append((i+1)+" ");
}
sb.append("\n");
//sSystem.out.println();
int arr[]= new int[n];
for(int k=0;k<n;k++)
{
arr[k]=k+1;
}
for(int i=1;i<n;i++)
{
int temp=arr[i];
arr[i]=arr[i-1];
arr[i-1]=temp;
for(int j=0;j<n;j++)
{
sb.append((arr[j])+" ");
}
sb.append("\n");
}
System.out.print(sb);
}
sc.close();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 60501ad7531e4f652ee5be79c27c74b3 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public final class Main{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static void print(int[] arr){
PrintWriter out=new PrintWriter(System.out);
for(int i=0;i<arr.length-1;i++) {
out.print(arr[i]+" ");
}
out.println(arr[arr.length-1]);
out.flush();
}
public static void main(String[] args) throws java.lang.Exception{
PrintWriter out=new PrintWriter(System.out);
FastReader sc = new FastReader();
int t=sc.nextInt();
// HashMap<Integer,ArrayList<int[]>> map = new HashMap<>();
// int[] dp = new int[]
while(t-->0) {
int n =sc.nextInt();
// if (map.containsKey(n)){
// System.out.println(n);
// for(int[] ppp:map.get(n)) {
// print(ppp);
// }
// continue;
// }
int[] arr = new int[n];
for(int i=1;i<=n;i++) {
arr[i-1]=i;
}
System.out.println(n);
// ArrayList<int[]> pp = new ArrayList<>();
// pp.add(arr.clone());
print(arr);
for(int i=1;i<n;i++) {
int tt=arr[i];
arr[i]=arr[i-1];
arr[i-1]=tt;
print(arr);
// pp.add(arr.clone());
}
// map.put(n, pp);
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 0efdb00d4e9b0f1130cbf47e4e41a217 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.io.*;
import java.math.*;
import java.util.*;
public class B_Permutation_Chain
{
public static void main(String[] args)throws Exception
{
new Solver().solve();
}
}
//* Success is not final, failure is not fatal: it is the courage to continue that counts.
class Solver {
void solve() throws Exception
{
int t = sc.nextInt();
while(t-->0){
int n =sc.nextInt();
// int x = n-1;
System.out.println(n);
// int r = n;
// for(int i = 1;i<=n;i++){
// System.out.print(i+" ");
// }
// System.out.println();
for(int k = n;k>=1;k--){
StringBuilder SB = new StringBuilder();
int i = 1;
while(i<k){
SB.append(i+" ");
i++;
}
SB.append(n+" ");
while(i<n){
SB.append(i+" ");
i++;
}
System.out.println(SB.toString());
}
}
sc.flush();
}
final Helper sc;
final int MAXN = 1000_006;
final long MOD = (long) 1e9 + 7;
Solver() {
sc = new Helper(MOD, MAXN);
sc.initIO(System.in, System.out);
}
}
class Helper {
final long MOD;
final int MAXN;
final Random rnd;
public Helper(long mod, int maxn) {
MOD = mod;
MAXN = maxn;
rnd = new Random();
}
public static int[] sieve;
public static ArrayList<Integer> primes;
public void setSieve() {
primes = new ArrayList<>();
sieve = new int[MAXN];
int i, j;
for (i = 2; i*i < MAXN; ++i)
if (sieve[i] == 0) {
primes.add(i);
for (j = i*i; j < MAXN; j += i) {
sieve[j] = i;
}
}
}
public static long[] factorial;
public void setFactorial() {
factorial = new long[MAXN];
factorial[0] = 1;
for (int i = 1; i < MAXN; ++i) factorial[i] = factorial[i - 1] * i % MOD;
}
public long getFactorial(int n) {
if (factorial == null) setFactorial();
return factorial[n];
}
public long ncr(int n, int r) {
if (r > n) return 0;
if (factorial == null) setFactorial();
long numerator = factorial[n];
long denominator = factorial[r] * factorial[n - r] % MOD;
return numerator * pow(denominator, MOD - 2, MOD) % MOD;
}
public long[] getLongArray(int size) throws Exception {
long[] ar = new long[size];
for (int i = 0; i < size; ++i) ar[i] = nextLong();
return ar;
}
public int[] getIntArray(int size) throws Exception {
int[] ar = new int[size];
for (int i = 0; i < size; ++i) ar[i] = nextInt();
return ar;
}
public long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
public int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
public long max(long[] ar) {
long ret = ar[0];
for (long itr : ar) ret = Math.max(ret, itr);
return ret;
}
public int max(int[] ar) {
int ret = ar[0];
for (int itr : ar) ret = Math.max(ret, itr);
return ret;
}
public long min(long[] ar) {
long ret = ar[0];
for (long itr : ar) ret = Math.min(ret, itr);
return ret;
}
public int min(int[] ar) {
int ret = ar[0];
for (int itr : ar) ret = Math.min(ret, itr);
return ret;
}
public long sum(long[] ar) {
long sum = 0;
for (long itr : ar) sum += itr;
return sum;
}
public long sum(int[] ar) {
long sum = 0;
for (int itr : ar) sum += itr;
return sum;
}
public void shuffle(int[] ar) {
int r;
for (int i = 0; i < ar.length; ++i) {
r = rnd.nextInt(ar.length);
if (r != i) {
ar[i] ^= ar[r];
ar[r] ^= ar[i];
ar[i] ^= ar[r];
}
}
}
public void shuffle(long[] ar) {
int r;
for (int i = 0; i < ar.length; ++i) {
r = rnd.nextInt(ar.length);
if (r != i) {
ar[i] ^= ar[r];
ar[r] ^= ar[i];
ar[i] ^= ar[r];
}
}
}
public long pow(long base, long exp, long MOD) {
base %= MOD;
long ret = 1;
while (exp > 0) {
if ((exp & 1) == 1) ret = ret * base % MOD;
base = base * base % MOD;
exp >>= 1;
}
return ret;
}
static byte[] buf = new byte[1000_006];
static int index, total;
static InputStream in;
static BufferedWriter bw;
public void initIO(InputStream is, OutputStream os) {
try {
in = is;
bw = new BufferedWriter(new OutputStreamWriter(os));
} catch (Exception e) {
}
}
public void initIO(String inputFile, String outputFile) {
try {
in = new FileInputStream(inputFile);
bw = new BufferedWriter(new OutputStreamWriter(
new FileOutputStream(outputFile)));
} catch (Exception e) {
}
}
private int scan() throws Exception {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0)
return -1;
}
return buf[index++];
}
public String nextLine() throws Exception {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c >= 32; c = scan())
sb.append((char) c);
return sb.toString();
}
public String next() throws Exception {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan())
sb.append((char) c);
return sb.toString();
}
public int nextInt() throws Exception {
int c, val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+')
c = scan();
for (; c >= '0' && c <= '9'; c = scan())
val = (val << 3) + (val << 1) + (c & 15);
return neg ? -val : val;
}
public long nextLong() throws Exception {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+')
c = scan();
for (; c >= '0' && c <= '9'; c = scan())
val = (val << 3) + (val << 1) + (c & 15);
return neg ? -val : val;
}
public void print(Object a) throws Exception {
bw.write(a.toString());
}
public void printsp(Object a) throws Exception {
print(a);
print(" ");
}
public void println() throws Exception {
bw.write("\n");
}
public void printArray(int[] arr) throws Exception{
for(int i = 0;i<arr.length;i++){
print(arr[i]+ " ");
}
println();
}
public void println(Object a) throws Exception {
print(a);
println();
}
public void flush() throws Exception {
bw.flush();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | a351fb198a903adc2013829ebf7cd0ff | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | /*input
Prateek Singh
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.*;
import java.util.*;
public class codeforces {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String args[]) {
if (System.getProperty("ONLINE_JUDGE") == null) {
// Input is a file
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
} catch (Exception e) {
System.err.println("Error");
}
} else {
// Input is System.in
}
FastReader sc = new FastReader();
// Scanner sc = new Scanner(System.in);
//System.out.println(java.time.LocalTime.now());
StringBuilder sb = new StringBuilder();
int t = sc.nextInt();
while (t > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 1; i <= n; i++)arr[i - 1] = i;
sb.append(n + "\n");
print(arr, sb);
arr[0] = n;
arr[n - 1] = 1;
print(arr, sb);
for (int i = 3; i <= n; i++) {
int temp = arr[0];
arr[0] = arr[i - 2];
arr[i - 2] = temp;
print(arr, sb);
}
//sb.append(ans + "\n");
t--;
}
System.out.println(sb);
}
public static void print(int[] arr, StringBuilder sb) {
for (int i : arr)sb.append(i + " ");
sb.append("\n");
}
public static int bs(ArrayList<Long> al, long tar) {
int l = 0;
int r = al.size() - 1;
while (l < r) {
int mid = (l + r) >> 1;
if (al.get(mid) < tar)l = mid + 1;
else r = mid;
}
if (al.get(l) <= tar )return l;
else return l - 1;
}
public static void dfs(Pair src, ArrayList<ArrayList<Pair>> al, ArrayList<Long> dist, ArrayList<Long> dist2, int[] ans) {
long a = src.a;
long b = src.b;
if (dist.size() != 0)a += dist.get(dist.size() - 1);
if (dist2.size() != 0)b += dist2.get(dist2.size() - 1);
dist.add(a);
dist2.add(b);
ans[src.des] = bs(dist2, dist.get(dist.size() - 1));
for (Pair p : al.get(src.des)) {
dfs(p, al, dist, dist2, ans);
}
dist.remove(dist.size() - 1);
dist2.remove(dist2.size() - 1);
}
public static boolean isPowerOfTwo(long x) {
return x != 0 && ((x & (x - 1)) == 0);
}
static class Pair {
int des;
long a;
long b;
Pair (int des, long a, long b) { //constructor
this.des = des;
this.a = a;
this.b = b;
}
}
//////////nCr////////////////////////////////////////
///////// SUM OF EACH DIGIT OF A NUMBER ///////////////
public static long digSum(long a) {
long sum = 0;
while (a > 0) {
sum += a % 10;
a /= 10;
}
return sum;
}
///////// TO CHECK NUMBER IS PRIME OR NOT ///////////////
public static boolean isPrime(int n) {
if (n <= 1)return false;
if (n <= 3)return true;
if (n % 2 == 0 || n % 3 == 0)return false;
for (int i = 5; i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0)return false;
}
return true;
}
///////// NEXT PRIME NUMBER BIGGER THAN GIVEN NUMBER ///////////////
public static int nextPrime(int n) {
while (true) {
n++;
if (isPrime(n)) break;
}
return n;
}
///////// GCD ///////////////
public static long gcd(long a, long b) {
if (b == 0)return a;
return gcd(b, a % b);
}
///////// LCM ///////////////
// public static int lcm(int a, int b) {
// return (a * b) / gcd(a, b);
// }
public static long factorial(int n) {
long ans = 1;
if (n == 0 || n == 1)return (ans);
for (int i = 2; i <= n; i++) {
ans *= i;
}
return ans;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 7a645c9eed20d8aa30ff80ef702c44e4 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static int h;
static int n;
static long mod = 1000000007;
public static void main(String[] args) {
Sca in = new Sca(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
out.println(n);
for (int i = 0; i < n; i++) {
out.printf("%d ", i + 1);
}
out.println();
for (int i = 1; i < n; i++) {
int j = 2;
for (; j < i + 2; j++) {
out.printf("%d ", j);
}
out.printf("1 ");
for (; j <= n; j++) {
out.printf("%d ", j);
}
out.println();
}
}
out.close();
}
static class Sca {
BufferedReader br;
StringTokenizer st;
public Sca(InputStream io) {
br = new BufferedReader(new InputStreamReader(io));
}
public String line() {
String result = "";
try {
result = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return result;
}
public String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public char[] nextc() {
return next().toCharArray();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 62e51ea93f1122f8312385f9eefa1151 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Solution127 {
private final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private final PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) {
Solution127 solution = new Solution127();
solution.solve();
solution.close();
}
private void solve() {
int t = readInt();
for (int i = 0; i < t; i++) {
solveCase();
}
}
void append(int[] a, StringBuilder sb) {
sb.append(a[0]);
for (int i = 1; i < a.length; i++) {
sb.append(" ");
sb.append(a[i]);
}
sb.append("\n");
}
private void solveCase() {
int n = readInt();
int[] a = new int[n];
for (int i = 1; i <= n; i++) {
a[i - 1] = i;
}
StringBuilder sb = new StringBuilder();
append(a, sb);
int t = a[0];
a[0] = a[1];
a[1] = t;
int l = 2;
append(a, sb);
l: while (true) {
for (int j = 0; j < n; j++) {
if (a[j] == j + 1) {
l++;
t = a[0];
a[0] = a[j];
a[j] = t;
append(a, sb);
continue l;
}
}
break;
}
pw.println(l + "\n" + sb.toString().trim());
}
private void close() {
pw.close();
}
private String readLine() {
String line = null;
try {
line = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return line;
}
private String readString() {
return readLine();
}
private long readLong() {
return Long.parseLong(readLine());
}
private int readInt() {
return Integer.parseInt(readLine());
}
private String[] stringArray() {
StringTokenizer st = new StringTokenizer(readLine());
int n = st.countTokens();
String ret[] = new String[n];
for (int i = 0; i < n; i++) {
ret[i] = st.nextToken();
}
return ret;
}
private int[] readIntArr() {
String[] str = stringArray();
int arr[] = new int[str.length];
for (int i = 0; i < arr.length; i++)
arr[i] = Integer.parseInt(str[i]);
return arr;
}
private double[] readDoubleArr() {
String[] str = stringArray();
double arr[] = new double[str.length];
for (int i = 0; i < arr.length; i++)
arr[i] = Double.parseDouble(str[i]);
return arr;
}
private long[] readLongArr() {
String[] str = stringArray();
long arr[] = new long[str.length];
for (int i = 0; i < arr.length; i++)
arr[i] = Long.parseLong(str[i]);
return arr;
}
private double readDouble() {
return Double.parseDouble(readLine());
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | e51c000373d3c67b8e0c699380dc63b4 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.InputStreamReader;
import java.lang.invoke.MethodHandles;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
public class c{
static int []f=new int [2000001];
//static int []a=new int [500001];
static long max=123456789;
public static void main(String []args) {
MyScanner s=new MyScanner();
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
int t=s.nextInt();
while(t-->0) {
int n=s.nextInt();
int []f=new int [n];
for(int i=0;i<n;i++)
f[i]=i+1;
int l=0;
System.out.println(n);
for(int x:f)
out.print(x+" ");
out.println();
while(l<n-1) {
int a=f[l];
f[l]=f[l+1];
f[l+1]=a;
for(int x:f)
out.print(x+" ");
out.println();
l++;
}
out.flush();
}
}
public static void dfs(int n,long sum,int u,int k) {
if(k>=n) {
if(k==n)
max=Math.min(sum,max);
return ;
}
for(int i=u;i<=9;i++) {
dfs(n,sum*10+i,i+1,k+i);
}
}
/*
public static void buildertree(int k,int l,int r) {
if(l==r)
{
f[k]=a[l];
return ;
}
int m=l+r>>1;
buildertree(k+k,l,m);
buildertree(k+k+1,m+1,r);
f[k]=
}
public static void update(int u,int l,int r,int x,int c)
{
if(l==x && r==x)
{
f[u]=c;
return;
}
int mid=l+r>>1;
if(x<=mid)update(u<<1,l,mid,x,c);
else if(x>=mid+1)update(u<<1|1,mid+1,r,x,c);
f[u]=Math.max(f[u+u], f[u+u+1]);
}
public static int query(int k,int l,int r,int x,int y) {
if(x==l&&y==r) {
return f[k];
}
int m=l+r>>1;
if(y<=m) {
return query(k+k,l,m,x,y);
}
else if(x>m)return query(k+k+1,m+1,r,x,y);
else {
int i=query(k+k,l,m,x,m),j=query(k+k+1,m+1,r,m+1,y);
return Math.max(j, Math.max(i+j, i));
}
}
public static void calc(int k,int l,int r,int x,int z) {
f[k]+=z;
if(l==r) {
return ;
}
int m=l+r>>1;
if(x<=m)
calc(k+k,l,m,x,z);
else calc(k+k+1,m+1,r,x,z);
}
*/
static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
try {
final String USERDIR = System.getProperty("user.dir");
String cname = MethodHandles.lookup().lookupClass().getCanonicalName().replace(".MyScanner", "");
cname = cname.lastIndexOf('.') > 0 ? cname.substring(cname.lastIndexOf('.') + 1) : cname;
final File fin = new File(USERDIR + "/io/c" + cname.substring(1,5) + "/" + cname + ".in");
br = new BufferedReader(new InputStreamReader(fin.exists()
? new FileInputStream(fin) : System.in));
} catch (Exception e) {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
public String next() {
try {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
} catch (Exception e) {
throw new RuntimeException(e);
}
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | a0fca170e78ea0f6cb2d2da71d8b1de3 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | /******************************************************************************
No one else matters, this is the moment, this is day.
*******************************************************************************/
import java.io.*;
import java.util.*;
public class Main {
static PrintWriter out=new PrintWriter(System.out);
public static void printArr(int arr[], int N){
for(int i=0; i<N; i++){
out.print(arr[i]+" ");
}
out.println("");
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
Scanner sc = new Scanner(System.in);
// int T = sc.nextInt();
int T = Integer.parseInt(br.readLine());
while (T-- > 0) {
int N = Integer.parseInt(br.readLine());
out.println(N);
int arr[] = new int[N];
for(int i=1; i<=N; i++) arr[i-1] = i;
printArr(arr, N);
for(int i=1; i<N; i++){
int t = arr[N-i-1];
arr[N-i-1] = arr[N-i];
arr[N-i] = t;
printArr(arr, N);
}
}
out.flush();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 7ba995590a19ad7bdb13acf40d3faff1 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
int pos = 1;
System.out.println(n);
for (int i = 1; i <= n; i++) {
StringBuilder str = new StringBuilder();
for (int j = 2; j <= i; j++) {
str.append(j + " ");
}
str.append(1 + " ");
for (int j = i + 1; j <= n; j++)
str.append(j+" ");
System.out.println(str);
}
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | e30c0b7358d07e8d93f65331ff018d31 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | //package codeforces;
import java.util.*;
import java.io.*;
public class CF {
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int q = Integer.parseInt(br.readLine());
while(q-->0) {
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
pw.println(n);
int[] arr = new int[n];
for (int i=0; i<n; i++) {
arr[i] = i+1;
pw.print(arr[i]+" ");
}
pw.println();
for (int i=0; i<n-1; i++) {
int t = arr[i];
arr[i] = arr[i+1];
arr[i+1] = t;
for (int j: arr) {
pw.print(j+" ");
}
pw.println();
}
}
pw.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 46bb85357dbed6952cd512ee4af21a49 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
/*
*/
public class B{
static FastReader sc=null;
public static void main(String[] args) {
sc=new FastReader();
int t=sc.nextInt();
PrintWriter out=new PrintWriter(System.out);
for(int tt=0;tt<t;tt++) {
int n=sc.nextInt();
out.println(n);
for(int i=0;i<n;i++) {
int ans[]=new int[n];
for(int j=0;j<n;j++)ans[j]=j+1;
for(int j=0;j<i;j++) ans[j]=ans[j+1];
ans[i]=1;
for(int e:ans)out.print(e+" ");
out.println();
}
}
out.close();
}
static int[] ruffleSort(int a[]) {
ArrayList<Integer> al=new ArrayList<>();
for(int i:a)al.add(i);
Collections.sort(al);
for(int i=0;i<a.length;i++)a[i]=al.get(i);
return a;
}
static void print(int a[]) {
for(int e:a) {
System.out.print(e+" ");
}
System.out.println();
}
static class FastReader{
StringTokenizer st=new StringTokenizer("");
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String next() {
while(!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
}
catch(IOException e){
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
int[] readArray(int n) {
int a[]=new int[n];
for(int i=0;i<n;i++)a[i]=sc.nextInt();
return a;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | efd4dc17ae451699f10eb4e793bd7d80 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | // package com.company.Codeforces;/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class newPro
{
public static void solve(){
int n = sc.nextInt();
int [] p = new int [n];
for(int i=0; i<n; i++){
p[i] = i+1;
}
out.println(n);
for(int i=0; i<n; i++){
for(int num : p){
out.print(num + " ");
}
out.println();
if(i < n-1){
int tmp = p[i];
p[i] = p[i+1];
p[i+1] = tmp;
}
}
}
static PrintWriter out;
static Scanner sc;
public static void main (String[] args) throws java.lang.Exception
{
out = new PrintWriter(System.out);
sc = new Scanner();
int t = sc.nextInt();
while(t-->0){
solve();
}
out.flush();
out.close();
}
private static long power(long a, long b) {
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
private static long power(long a, long b, long m) {
a %= m;
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = res * a % m;
a = a * a % m;
b >>= 1;
}
return res;
}
private static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
private static long gcd(long a, long b) {
return (b == 0) ? a : gcd(b, a % b);
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() {
br = new BufferedReader(new
InputStreamReader(System.in));
/*
try {
br = new BufferedReader(new
InputStreamReader(new FileInputStream(new File("file_i_o\\input.txt"))));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
*/
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
Integer[] nextIntegerArray(int n) {
Integer[] array = new Integer[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; i++) {
array[i] = nextLong();
}
return array;
}
String[] nextStringArray() {
return nextLine().split(" ");
}
String[] nextStringArray(int n) {
String[] array = new String[n];
for (int i = 0; i < n; i++) {
array[i] = next();
}
return array;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 42276c066f6535b51961fa41537cee9e | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | //Code By KB.
import java.beans.Visibility;
import java.io.*;
import java.lang.annotation.Target;
import java.lang.reflect.Array;
import java.nio.channels.AsynchronousCloseException;
import java.security.KeyStore.Entry;
import java.util.*;
import java.util.logging.*;
import java.io.*;
import java.util.logging.*;
import java.util.regex.*;
import javax.management.ValueExp;
import javax.print.DocFlavor.INPUT_STREAM;
import javax.swing.plaf.basic.BasicBorders.SplitPaneBorder;
import javax.swing.text.AbstractDocument.LeafElement;
import javax.xml.validation.Validator;
public class Codeforces {
public static void main(String[] args) {
FlashFastReader in = new FlashFastReader(System.in);
try(PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));)
{
new Codeforces().solution(in, out);
} catch (Exception e) {
System.out.println(e.getStackTrace());
}
}
public void solution(FlashFastReader in, PrintWriter out)
{
try {
int t = in.nextInt();
while (t-->0) {
int n = in .nextInt();
int a[] = new int[n];
out.println(n);
for (int i = 0; i < n ; i++) {
a[i] = i+1;
out.print(a[i]+" ");
}
out.println();
for (int i = 0; i < a.length-1; i++) {
int temp= a[i];
a[i]=a[i+1];
a[i+1]=temp;
for (int j = 0; j < a.length; j++) {
out.print(a[j]+" ");
}
out.println();
}
}
} catch (Exception exception) {
out.println(exception);
}
}
public int countSquares(int[][] matrix) {
int dp [][] = new int [matrix.length+1][matrix[0].length+1];
for (int i = 1; i <=matrix.length; i++) {
for (int j = 1; j <=matrix[0].length; j++) {
if (matrix[i][j]==1) {
if (i>=1&&j>=1) {
dp[i][j] =1+ Math.min(dp[i-1][j], Math.min(dp[i-1][j-1], dp[i][j-1]) );
}
}
x+=dp[i][j];
}
}
return x;
}
HashMap<Integer, Integer> map = new HashMap<>();
public int totalNQueens(int n) {
if (n==1) {
return 1;
}
if (n==2||n==3) {
return 0;
}
dfsNqueens( map , 0 , n );
return x;
}
public void dfsNqueens( HashMap<Integer, Integer>map , int col , int n ) {
if (col == n) {
x++;
return;
}
for (int i = 0; i < n; i++) {
map.put(col, i);
boolean f = true;
for (int j = 0; j < n; j++) {
if (map.containsKey(j)) {
if (map.get(j)==i||Math.abs(col - j)==Math.abs(map.get(j)-i) ) {
f = false;
}
}
}
if (f) {
dfsNqueens(map, col+1, n);
}
}
}
public int[] minOperations(String boxes) {
int a [] = new int[boxes.length()];
for (int i = 0; i < boxes.length(); i++) {
StringBuilder sb = new StringBuilder(boxes);
int x = 0;
for (int j = 0; j < sb.length(); j++) {
if (sb.indexOf("1")<0) {
break;
}
x+=Math.abs(i - sb.indexOf("1"));
sb.setCharAt(sb.indexOf("1"), '0');
}
a[i] = x;
}
return a;
}
public List<List<Integer>> subsets(int[] nums) {
List<Integer> curr = new ArrayList<>();
subsetsGen(0, nums , curr);
return list;
}
public void subsetsGen(int j ,int[] n , List<Integer>curr) {
list.add(curr);
for (int i = j; i < n.length; i++) {
curr.add(n[i]);
subsetsGen(j+1, n, curr);
curr.remove(curr.size()-1);
}
}
boolean visited[]= new boolean[1000];
public int findCircleNum(int[][] isConnected) {
for (int i = 0; i < isConnected.length; i++) {
if (visited[i]!=true) {
x++;
dfsFindProvinces( i , isConnected );
}
}
return x;
}
public void dfsFindProvinces( int i ,int [][]isConnected ) {
visited[i]=true;
for (int j = 0; j < isConnected.length; j++) {
if (isConnected[i][j]==1&&visited[j]!=true) {
dfsFindProvinces(j, isConnected);
}
}
}
public int[] canSeePersonsCount(int[] heights) {
int ans [] = new int[heights.length];
Stack<Integer> stck = new Stack<>();
for (int i = 0; i < heights.length; i++) {
while(!stck.isEmpty()&&heights[stck.peek()]<heights[i]) {
ans[stck.pop()]++;
}
if (!stck.isEmpty()) {
ans[stck.peek()]++;
}
stck.push(i);
}
return ans;
}
public int pathSum(TreeNode root, int targetSum) {
List<Integer> path = new ArrayList<>();
dsfPathSum(root, targetSum , 0 ,path);
pathSum(root.left, targetSum);
pathSum(root.right, targetSum);
for (List<Integer> p : list) {
for (int i = 0; i < p.size() ; i++) {
System.out.print(p.get(i)+" ");
}
System.out.println();
}
return list.size();
}
public void dsfPathSum(TreeNode root , int t , long sum , List<Integer>path ) {
if (root==null) {
return;
}
sum+=root.val ;
path.add(root.val);
if (sum==t) {
list.add(path);
}
dsfPathSum(root.left, t, sum , path);
dsfPathSum(root.right, t, sum , path);
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode l = new ListNode();
int a = 0;
int b =0;
l1 = reverseList(l1);
l2 = reverseList(l2);
while (l1!=null) {
a=a*10+l1.val;
l1=l1.next;
}
while (l2!=null) {
b=b*10+l2.val;
l2=l2.next;
}
a=a+b;
ListNode head = null;
while (a!=0) {
int r = a%10;
if(l==null) {
head= new ListNode(r);
l=head;
}else {
l.next = new ListNode(r);
l= l.next;
}
a/=10;
}
return head;
}
public ListNode reverseList(ListNode head) {
if(head==null || head.next==null)
return head;
ListNode nextNode=head.next;
ListNode newHead=reverseList(nextNode);
nextNode.next=head;
head.next=null;
return newHead;
}
int x=0;
public int uniquePathsIII(int[][] grid) {
int zeros = 0;
int sx=0;
int sy =0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j]==0) {
zeros++;
} else if (grid[i][j]==1) {
sx=i;
sy=j;
}
}
}
dfsUniquePaths(zeros , sx , sy , grid);
return x;
}
public void dfsUniquePaths(int zeros ,int i ,int j , int g[][]) {
if (i<0||i>=g.length||j>=g[0].length||j<0||g[i][j]<0) {
System.out.println(i+" "+j);
return;
}
if (g[i][j]==2) {
if(zeros==0)
x++;
return;
}
g[i][j]=-2;
zeros--;
dfsUniquePaths(zeros, i+1, j, g);
dfsUniquePaths(zeros, i-1, j, g);
dfsUniquePaths(zeros, i, j+1, g);
dfsUniquePaths(zeros, i, j-1, g);
g[i][j]=0;
zeros++;
}
public TreeNode addOneRow(TreeNode root, int val, int depth) {
if (depth==1) {
TreeNode t = new TreeNode(val);
t.left=root;
return t;
}
dfsAddOneRow(root , 1 , val , depth);
return root;
}
public void dfsAddOneRow(TreeNode root , int currD , int val , int depth) {
if (root==null) {
return ;
}
if (currD==depth) {
TreeNode l = root.left;
root.left = new TreeNode(val);
root.left.left = l;
TreeNode r = root.right;
root.right = new TreeNode(val);
root.right.right = r;
}
currD++;
dfsAddOneRow(root.left, currD, val, depth);
dfsAddOneRow(root.right, currD, val, depth);
}
public List<List<Integer>> threeSum(int[] nums) {
int n = nums.length;
int sum =0;
HashSet<List<Integer>> h = new HashSet<List<Integer>>();
for (int i = 0; i < nums.length; i++) {
sum=nums[i];
for (int j = i+1; j < nums.length; j++) {
sum+=nums[j];
int k =j;
n = nums.length;
while (k<n) {
int mid = (k+n)/2;
if ((sum+nums[mid])==0) {
HashSet<Integer> m = new HashSet<>();
m.add(nums[i]);
m.add(nums[j]);
m.add(nums[mid]);
if (m.size()==3) {
List<Integer>x = new ArrayList<>();
x.add(nums[i]);
x.add(nums[j]);
x.add( nums[mid]);
h.add(x);
}
} else if ((sum+nums[mid])>0) {
n=mid;
} else {
k=mid;
}
}
}
}
return list;
}
public int uniquePaths(int m, int n) {
int dp[][]= new int [m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if(i==0&&j==0) {
dp[i][j]=1;
} else if (i==0) {
dp[i][j]=dp[i][j]+dp[i][j-1];
} else if (j==0) {
dp[i][j]=dp[i][j]+dp[i-1][j];
} else {
dp[i][j]+=dp[i][j-1]+dp[i-1][j];
}
}
}
return dp[m-1][n-1];
}
public int maxProfit(int[] prices) {
int p = -1;
return p ;
}
public int trap(int[] height) {
int n = height.length;
int left[] = new int [n];
int right[] = new int [n];
left[0] = height[0];
for (int i = 1; i <n; i++) {
left[i] = Math.max(height[i], left[i-1]);
}
right[n-1]=height[n-1];
for (int i = n-2; i>=0 ; i--) {
right[i] = Math.max(height[i], right[i+1]);
}
for (int i = 1; i < right.length-1; i++) {
int s =Math.min(left[i] , right[i])-height[i];
if (s>0) {
x+=s;
}
}
return x;
}
public int maxProduct(int[] nums) {
int maxSoFar = Integer.MIN_VALUE;
int maxTillEnd = 0;
for (int i = 0; i < nums.length; i++) {
maxTillEnd*=nums[i];
if (maxSoFar<maxTillEnd) {
maxSoFar=maxTillEnd;
}
if (maxTillEnd<=0) {
maxTillEnd=1;
}
}
return maxSoFar;
}
public int maximumScore(int[] nums, int[] multipliers) {
int n = multipliers.length;
int dp[] = new int[n+1];
for (int i = 1; i < multipliers.length; i++) {
for (int j = 1; j < nums.length; j++) {
dp[i] = Math.max(dp[j-1], multipliers[i-1]*nums[i-1]);
}
}
return dp[n];
}
public int islandPerimeter(int[][] grid) {
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if(grid[i][j]==1){
dfsIslandPerimeter(i , j , grid);
}
}
}
return x;
}
public void dfsIslandPerimeter(int i , int j , int[][] grid) {
if(i<0||i>=grid.length||j<0||j>grid[0].length) {
x++;
return;
}
if(grid[i][j]==0) {
x++;
return;
}
if(grid[i][j]==1) {
return;
}
dfsIslandPerimeter(i+1, j, grid);
dfsIslandPerimeter(i-1, j, grid);
dfsIslandPerimeter(i, j+1, grid);
dfsIslandPerimeter(i, j-1, grid);
}
public int[] findOriginalArray(int[] changed) {
int n =changed.length;
int m = n/2;
int a[] = new int[m];
if (n<2) {
return new int[]{};
}
if (n%2!=0) {
return new int[]{};
}
TreeMap<Integer , Integer> map = new TreeMap<>();
for (int i = 0; i < changed.length; i++) {
map.put(changed[i], map.getOrDefault(changed[i] ,0)+1);
}
int j =0;
for (Map.Entry<Integer , Integer> e : map.entrySet()) {
if (map.containsKey(e.getKey()*2)&&map.get(e.getKey()*2)>=1&&map.get(e.getKey())>=1) {
map.put(e.getKey(), map.get(e.getKey()) -1);
if(j==n-1)
break;
a[j++] = e.getKey();
}
}
return a;
}
public int jump(int[] nums) {
int n =nums.length;
int jumps =1;
int i =0;
int j =0;
while (i<n) {
int max =0;
for (j=i+1 ;j<=i+nums[i];j++) {
max = Math.max(max , nums[j]+j);
}
jumps++;
i=max;
}
return jumps;
}
static void sort(int[] arr) {
if (arr.length < 2) return;
int mid = arr.length / 2;
int[] left_half = new int[mid];
int[] right_half = new int[arr.length - mid];
// copying the elements of array into left_half
for (int i = 0; i < mid; i++) {
left_half[i] = arr[i];
}
// copying the elements of array into right_half
for (int i = mid; i < arr.length; i++) {
right_half[i - mid] = arr[i];
}
sort(left_half);
sort(right_half);
merge(arr, left_half, right_half);
}
static void merge(int[] arr, int[] left_half, int[] right_half) {
int i = 0, j = 0, k = 0;
while (i < left_half.length && j < right_half.length) {
if (left_half[i] < right_half[j]) {
arr[k++] = left_half[i++];
}
else {
arr[k++] = right_half[j++];
}
}
while (i < left_half.length) {
arr[k++] = left_half[i++];
}
while (j < right_half.length) {
arr[k++] = right_half[j++];
}
}
public int minCostClimbingStairs(int[] cost) {
int n = cost.length;
int dp[] = new int[n+1];
dp[1] = cost[0];
dp[2] = cost[1];
for (int i = 3; i <=n; i++) {
dp[i]=cost[i-1]+Math.min(dp[i-1], dp[i-2]);
}
return dp[n];
}
TreeNode newAns =null;
public TreeNode removeLeafNodes(TreeNode root, int target) {
newAns = root;
dfsRemoveNode(newAns , target);
return newAns;
}
public void dfsRemoveNode(TreeNode root , int t) {
if (root==null) {
return;
}
if (root.val==t) {
if (root.left!=null) {
root=root.left;
}else if (root.right!=null){
root=root.right;
} else {
root=null;
}
}
dfsRemoveNode(root.left, t);
dfsRemoveNode(root.right, t);
}
public int maxProfit(int k, int[] prices) {
int n = prices.length;
if (n <= 1)
return 0;
if (k >= n/2) {
return stocks(prices, n);
}
int[][] dp = new int[k+1][n];
for (int i = 1; i <=k; i++) {
int c= dp[i-1][0] - prices[0];
for (int j = 1; j < n; j++) {
dp[i][j] = Math.max(dp[i][j-1], prices[j] + c);
c = Math.max(c, dp[i-1][j] - prices[j]);
}
}
return dp[k][n-1];
}
public int stocks(int [] prices , int n ) {
int maxPro = 0;
for (int i = 1; i < n; i++) {
if (prices[i] > prices[i-1])
maxPro += prices[i] - prices[i-1];
}
return maxPro;
}
public List<List<Integer>> combine(int n, int k) {
List<Integer> x = new ArrayList<>();
combinations( n , k , 1 ,x);
return list;
}
public void combinations(int n , int k ,int startPos , List<Integer> x) {
if (k==x.size()) {
list.add(new ArrayList<>(x));
//x=new ArrayList<>();
return;
}
for (int i = startPos; i <=n; i++) {
x.add(i);
combinations(n, k, i+1, x);
x.remove(x.size()-1);
}
}
List<List<Integer>> list = new ArrayList<>();
public List<List<Integer>> permute(int[] a) {
List<Integer> x = new ArrayList<>();
permutations(a , 0 ,a.length ,x );
return list ;
}
public void permutations(int a[] , int startPos , int l , List<Integer>x) {
if (l==x.size()) {
list.add(new ArrayList<>(x));
//x=new ArrayList<>();
}
for (int i = 0; i <=l; i++) {
if (!x.contains(a[i])) {
x.add(a[i]);
permutations(a, i, l, x);
x.remove(x.size()-1);
}
}
}
List<String> str = new ArrayList<>();
public List<String> letterCasePermutation(String s) {
casePermutations(s , 0 , s.length() , "" );
return str;
}
public void casePermutations(String s ,int i , int l , String curr) {
if (curr.length()==l) {
str.add(curr);
return;
}
char b = s.charAt(i);
curr+=b;
casePermutations(s, i+1, l, curr);
curr = new StringBuilder(curr).deleteCharAt(curr.length()-1).toString();
if (Character.isAlphabetic(b)&&Character.isLowerCase(b)) {
curr+=Character.toUpperCase(b);
casePermutations(s, i+1, l, curr);
curr = new StringBuilder(curr).deleteCharAt(curr.length()-1).toString();
} else if (Character.isAlphabetic(b)) {
curr+=Character.toLowerCase(b);
casePermutations(s, i+1, l, curr);
curr = new StringBuilder(curr).deleteCharAt(curr.length()-1).toString();
}
}
TreeNode ans= null;
public TreeNode lcaDeepestLeaves(TreeNode root) {
dfslcaDeepestLeaves(root , 0 );
return ans;
}
public void dfslcaDeepestLeaves(TreeNode root , int level) {
if (root==null) {
return;
}
if (depth(root.left)==depth(root.right)) {
ans= root;
return;
} else if (depth(root.left)>depth(root.right)){
dfslcaDeepestLeaves(root.left, level+1);
} else {
dfslcaDeepestLeaves(root.right, level+1);
}
}
public int depth(TreeNode root) {
if (root==null) {
return 0;
}
return 1+Math.max(depth(root.left), depth(root.right));
}
TreeMap<Integer , Integer> m = new TreeMap<>();
public int maxLevelSum(TreeNode root) {
int maxlevel =0;
int mx = Integer.MIN_VALUE;
dfsMaxLevelSum(root , 0);
for (Map.Entry<Integer,Integer> e : m.entrySet()) {
if (e.getValue()>mx) {
mx=e.getValue();
maxlevel=e.getKey()+1;
}
}
return maxlevel;
}
public void dfsMaxLevelSum(TreeNode root , int currLevel) {
if (root==null) {
return;
}
if (!m.containsKey(currLevel)) {
m.put(currLevel, root.val);
} else {
m.put(currLevel, m.get(currLevel)+root.val);
}
dfsMaxLevelSum(root.left, currLevel+1);
dfsMaxLevelSum(root.right, currLevel+1);
}
int teampPerf = 0;
int teampSpeed = 0;
public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
int [][] map = new int[efficiency.length][2];
for (int i = 0; i < efficiency.length; i++) {
map[i][0] = efficiency[i];
map[i][1] = speed[i];
}
Arrays.sort(map , (e1 , e2 )-> (e2[0] - e1[0]));
PriorityQueue<Integer> pq = new PriorityQueue<>();
calmax(map , speed , efficiency , pq , k);
return teampPerf ;
}
public void calmax(int [][]map , int s[] , int e[] , PriorityQueue<Integer>pq , int k) {
for (int i = 0 ; i<n ; i++) {
if (pq.size()==k) {
int lowestSpeed =pq.remove();
teampSpeed-=lowestSpeed;
}
pq.add(map[i][1]);
teampSpeed+=map[i][1];
teampPerf=Math.max(teampPerf, teampSpeed*map[i][1]);
}
}
int maxRob = 0;
public int rob(int[] nums) {
int one = 0;
int two = 0;
for (int i = 0; i < nums.length; i++) {
if (i%2==0) {
one+=nums[i];
} else {
two+=nums[i];
}
}
return Math.max(one, two);
}
public int numberOfWeakCharacters(int[][] properties) {
int c =0;
Arrays.sort(properties, (a, b) -> (b[0] == a[0]) ? (a[1] - b[1]) : b[0] - a[0]);
PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
pq.offer(0);
for (int i = 0; i < properties.length; i++) {
if (properties[i][1]<pq.peek()) {
c++;
}
pq.offer(properties[i][1]);
}
return c;
}
public boolean canFinish(int numCourses, int[][] prerequisites) {
boolean f = true ;
TreeMap <Integer,Integer> map = new TreeMap<>();
for (int i = 0; i < prerequisites.length; i++) {
if (map.get(prerequisites[i][1])!=null) {
return false;
}
map.put(prerequisites[i][0], prerequisites[i][1]);
}
return f;
}
int n =0;
char res[][] = new char [1000][1000];
public int countBattleships(char[][] board) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
res[i][j]= board[i][j];
}
}
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (res[i][j]=='X') {
int x = dfsB( i , j ,board , res);
n++;
}
}
}
return n ;
}
public int dfsB(int i , int j , char [][]b , char [][]res) {
if (i<0||i>=b.length|j<0||j>=b[0].length||res[i][j]=='.') {
return 0;
}
if (res[i][j]=='X') {
return 1+dfsB(i+1, j, b, res)+dfsB(i, j+1, b, res);
}
return 0;
}
List<Integer> l = new ArrayList<>();
public List<Integer> rightSideView(TreeNode root) {
return dfsRightSideView(root);
}
public List<Integer> dfsRightSideView(TreeNode root) {
if (root!=null) {
l.add(root.val);
}
if (root==null) {
return l;
}
if (root.right==null) {
l.add(root.right.val);
}
dfsRightSideView(root.right);
return l;
}
public void dfsSumNumber(TreeNode root ,int sum , int l ) {
l=l*10 +root.val;
if (root.left!=null) {
dfsSumNumber(root.left,sum , l);
}
if (root.right!=null) {
dfsSumNumber(root.right , sum, l);
}
if (root.left==null && root.right==null) {
sum+=l;
}
// r=l*10+root.val;
l-=root.val;
l/=10;
}
//BFS SOLN
public int[][] updateMatrix(int[][] mat) {
int m =mat.length;
int n = mat[0].length;
Queue<int[]> q = new LinkedList<>();
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat[0].length; j++) {
if (mat[i][j]==0) {
q.offer(new int[] { i , j});
} else {
mat[i][j] = Integer.MAX_VALUE;
}
}
}
int dir[][] = {{1 , 0} , {-1 , 0} , {0 , 1} , { 0 , -1}};
while (!q.isEmpty()) {
int zeroPos[] = q.poll();
for (int[] d : dir) {
int r = zeroPos[0]+d[0];
int c = zeroPos[1]+d[1];
if (r<0||r>=mat.length || c<0||c>=mat[0].length||mat[r][c]<=mat[zeroPos[0]][zeroPos[1]]+1) {
continue;
}
q.add(new int[]{r,c});
mat[r][c] = mat[zeroPos[0]][zeroPos[1]]+1;
}
}
return mat;
}
// DFS SOLN
// public int[][] updateMatrix(int[][] mat) {
// int res [][] = new int[mat.length][mat[0].length];
// for (int i = 0; i < mat.length; i++) {
// for (int j = 0; j < mat.length; j++) {
// if (mat[i][j]==0) {
// Dis0(i, j, mat , res, 0);
// }
// }
// }
// return res;
// }
// public void Dis0(int i , int j , int [][]mat, int res[][] , int dist) {
// if (i<0||i>=mat.length||j>=mat[0].length||j<0) {
// return ;
// }
// if (dist==0 ||mat[i][j]==1&&(res[i][j]==0||res[i][j]>dist)) {
// res[i][j]= dist;
// Dis0(i+1, j, mat, res, dist+1);
// Dis0(i-1, j, mat, res, dist+1);
// Dis0(i, j+1, mat, res, dist+1);
// Dis0(i, j-1, mat, res, dist+1);
// }
// // return dist;
// }
public int maxDepth(TreeNode root) {
if (root==null) {
return 0;
}
return maxD(root , 0);
}
public int maxD(TreeNode root , int d ) {
if (root==null) {
return d;
}
return Math.max(maxD(root.left, d+1), maxD(root.right, d+1));
}
public int reverse(int x) {
int sign=x>0?1:-1;
//x=Math.abs(x);
long s= 0;
int r= 0;
int n=x;
while (n!=0) {
r=n%10;
s=s*10+r;
n/=10;
if (s>Integer.MAX_VALUE||s<Integer.MIN_VALUE) {
return 0;
}
}
return (int)s*sign;
}
public boolean checkInclusion(String s1, String s2) {
boolean f = false ;
if (s1.length()>s2.length()) {
return false;
}
for (int i = 0; i < s2.length(); i++) {
HashMap<Character ,Integer> c1 = new HashMap<>();
HashMap<Character ,Integer> c2 = new HashMap<>();
for (int j = 0; j < s1.length(); j++) {
char ch2 = s2.charAt(i+j);
char ch1 = s1.charAt(j);
// if (c1.get(key)) {
// }
}
}
return f;
}
public class ListNode {
int val;
ListNode next;
ListNode() {}
ListNode(int val) { this.val = val; }
ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}
public ListNode middleNode(ListNode head) {
ListNode mid = head;
ListNode last = head;
int k =0;
while (last.next!=null&&last.next.next!=null) {
k++;
mid= mid.next;
last = last.next.next;
}
if (getLen(head)%2==0) {
return mid.next;
}
return mid;
}
public int getLen(ListNode mid) {
int l = 0;
ListNode p = mid;
while (p!=null) {
l++;
p=p.next;
}
return l;
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
public List<List<Integer>> verticalTraversal(TreeNode root) {
TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map = new TreeMap<>();
dfs(root, 0, 0, map);
List<List<Integer>> list = new ArrayList<>();
for (TreeMap<Integer, PriorityQueue<Integer>> ys : map.values()) {
list.add(new ArrayList<>());
for (PriorityQueue<Integer> nodes : ys.values()) {
while (!nodes.isEmpty()) {
// list.get(list.size() - 1).add(nodes.poll());
}
}
}
return list;
}
private void dfs(TreeNode root, int x, int y, TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map) {
if (root == null) {
return;
}
if (!map.containsKey(x)) {
map.put(x, new TreeMap<>());
}
if (!map.get(x).containsKey(y)) {
map.get(x).put(y, new PriorityQueue<>());
}
map.get(x).get(y).offer(root.val);
dfs(root.left, x - 1, y + 1, map);
dfs(root.right, x + 1, y + 1, map);
}
public long pow (int x ,int n ) {
long y = 1;
if (n==0) {
return y;
}
while (n>0) {
if (n%2!=0) {
y=y*x;
}
x*=x;
n/=2;
}
return y;
}
public long powrec (int x ,int n ) {
long y = 1;
if (n==0) {
return y;
}
y = powrec(x, n/2);
if (n%2==0) {
return y*y;
}
return y*y*x;
}
public int fib(int n) {
if (n==0||n==11) {
return n;
}
return fib(n-1)+fib(n-2);
}
public long gcd(long a , long b)
{
if (b%a==0) {
return a;
}
return gcd(b%a,a);
}
public int maximumElement(int a[])
{
int x = Integer.MIN_VALUE;
for (int i = 0; i < a.length; i++) {
if (a[i]>x) {
x=a[i];
}
}
return x;
}
public int minimumElement(int a[])
{
int x = Integer.MAX_VALUE;
for (int i = 0; i < a.length; i++) {
if (a[i]<x) {
x=a[i];
}
}
return x;
}
public boolean [] sieveOfEratosthenes(int n) {
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
if (prime[p]) {
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
// GFG arraylist function for referral
public ArrayList create2DArrayList(int n ,int k)
{
// Creating a 2D ArrayList of Integer type
ArrayList<ArrayList<Integer> > x
= new ArrayList<ArrayList<Integer> >();
// One space allocated for R0
// Adding 3 to R0 created above x(R0, C0)
//x.get(0).add(0, 3);
int xr = 0;
for (int i = 1; i <= n; i+=2) {
if ((i+k)*(i+1)%4==0) {
x.add(new ArrayList<Integer>());
x.get(xr).addAll(Arrays.asList(i , i+1));
}
else {
x.add(new ArrayList<Integer>());
x.get(xr).addAll(Arrays.asList(i+1 , i));
}
xr++;
}
// Creating R1 and adding values
// Note: Another way for adding values in 2D
// collections
// x.add(
// new ArrayList<Integer>(Arrays.asList(3, 4, 6)));
// // Adding 366 to x(R1, C0)
// x.get(1).add(0, 366);
// // Adding 576 to x(R1, C4)
// x.get(1).add(4, 576);
// // Now, adding values to R2
// x.add(2, new ArrayList<>(Arrays.asList(3, 84)));
// // Adding values to R3
// x.add(new ArrayList<Integer>(
// Arrays.asList(83, 6684, 776)));
// // Adding values to R4
// x.add(new ArrayList<>(Arrays.asList(8)));
// // Appending values to R4
// x.get(4).addAll(Arrays.asList(9, 10, 11));
// // Appending values to R1, but start appending from
// // C3
// x.get(1).addAll(3, Arrays.asList(22, 1000));
// This method will return 2D array
return x;
}
}
class FlashFastReader
{
BufferedReader in;
StringTokenizer token;
public FlashFastReader(InputStream ins)
{
in=new BufferedReader(new InputStreamReader(ins));
token=new StringTokenizer("");
}
public boolean hasNext()
{
while (!token.hasMoreTokens())
{
try
{
String s = in.readLine();
if (s == null) return false;
token = new StringTokenizer(s);
} catch (IOException e)
{
throw new InputMismatchException();
}
}
return true;
}
public String next()
{
hasNext();
return token.nextToken();
}
public int nextInt()
{
return Integer.parseInt(next());
}
public int[] nextIntsInputAsArray(int n)
{
int[] res = new int[n];
for (int i = 0; i < n; i++)
res[i] = nextInt();
return res;
}
public long nextLong() {
return Long.parseLong(next());
}
public long[] nextLongsInputAsArray(int n)
{
long [] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public String nextString() {
return String.valueOf(next());
}
public String[] nextStringsInputAsArray(int n)
{
String [] a = new String[n];
for (int i = 0; i < n; i++)
a[i] = nextString();
return a;
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | d6978590cb78d5889f5b06db04cf91f1 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.OutputStream;
import java.io.BufferedOutputStream;
import java.util.*;
public class CodeForces_Edu_133 {
public static void main(String[] args)throws Exception{
FastScanner in = new FastScanner();OutputStream out = new BufferedOutputStream ( System.out );
int t=in.nextInt();
while (t-->0){
int n=in.nextInt();int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=i+1;
}
out.write((n+"\n").getBytes());
for(int j=0;j<n;j++){
out.write((a[j]+" ").getBytes());
}
out.write("\n".getBytes());
for(int i=1;i<n;i++){
int temp=a[i];a[i]=a[i-1];a[i-1]=temp;
for(int j=0;j<n;j++){
out.write((a[j]+" ").getBytes());
}
out.write("\n".getBytes());
}
}
out.flush();
}
public static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(String s) {
try {
br = new BufferedReader(new FileReader(s));
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String nextToken() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(nextToken());
}
long nextLong() {
return Long.parseLong(nextToken());
}
double nextDouble() {
return Double.parseDouble(nextToken());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | c8b192c74d3b0d1562aed6d760c27333 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.StringTokenizer;
public class B {
static PrintWriter out = new PrintWriter(System.out);
static MyFastReaderB in = new MyFastReaderB();
static long mod = (long) (1e9 + 7);
public static void main(String[] args) throws Exception {
//Scanner sc=new Scanner(System.in);
int test = i();
while (test-- > 0) {
int n=i();
int[] arr=new int[n];
out.print(n+"\n");
for(int i=1;i<=n;i++) arr[i-1]=i;
for(int p: arr) out.print(p+" ");
out.print("\n");
for(int i=n-1;i>0;i--) {
int x=arr[i];
arr[i]=arr[i-1];
arr[i-1]=x;
for(int p: arr) out.print(p+" ");
out.print("\n");
}
out.flush();
}
out.close();
}
static class pair {
long x, y;
pair(long ar, long ar2) {
x = ar;
y = ar2;
}
}
static void sort(long[] a) // check for long
{
ArrayList<Long> l = new ArrayList<>();
for (long i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
public static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
static class DescendingComparator implements Comparator<Integer> {
public int compare(Integer a, Integer b) {
return b - a;
}
}
static class AscendingComparator implements Comparator<Integer> {
public int compare(Integer a, Integer b) {
return a - b;
}
}
static boolean isPalindrome(char X[]) {
int l = 0, r = X.length - 1;
while (l <= r) {
if (X[l] != X[r])
return false;
l++;
r--;
}
return true;
}
static long fact(long N) {
long num = 1L;
while (N >= 1) {
num = ((num % mod) * (N % mod)) % mod;
N--;
}
return num;
}
static long pow(long a, long b) {
long mod = 1000000007;
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0)
pow = (pow * x) % mod;
x = (x * x) % mod;
b /= 2;
}
return pow;
}
static long toggleBits(long x)// one's complement || Toggle bits
{
int n = (int) (Math.floor(Math.log(x) / Math.log(2))) + 1;
return ((1 << n) - 1) ^ x;
}
static int countBits(long a) {
return (int) (Math.log(a) / Math.log(2) + 1);
}
static boolean isPrime(long N) {
if (N <= 1)
return false;
if (N <= 3)
return true;
if (N % 2 == 0 || N % 3 == 0)
return false;
for (int i = 5; i * i <= N; i = i + 6)
if (N % i == 0 || N % (i + 2) == 0)
return false;
return true;
}
static long GCD(long a, long b) {
if (b == 0) {
return a;
} else
return GCD(b, a % b);
}
// Debugging Functions Starts
static void print(char A[]) {
for (char c : A)
System.out.print(c + " ");
System.out.println();
}
static void print(boolean A[]) {
for (boolean c : A)
System.out.print(c + " ");
System.out.println();
}
static void print(int A[]) {
for (int a : A)
System.out.print(a + " ");
System.out.println();
}
static void print(long A[]) {
for (long i : A)
System.out.print(i + " ");
System.out.println();
}
static void print(ArrayList<Integer> A) {
for (int a : A)
System.out.print(a + " ");
System.out.println();
}
// Debugging Functions END
// ----------------------
// IO FUNCTIONS STARTS
static HashMap<Integer, Integer> getHashMap(int A[]) {
HashMap<Integer, Integer> mp = new HashMap<>();
for (int a : A) {
int f = mp.getOrDefault(a, 0) + 1;
mp.put(a, f);
}
return mp;
}
public static Map<Character, Integer> mapSortByValue(Map<Character, Integer> hm) {
// Create a list from elements of HashMap
List<Map.Entry<Character, Integer>> list = new LinkedList<Map.Entry<Character, Integer>>(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Character, Integer>>() {
public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) {
return o1.getValue() - o2.getValue();
}
});
// put data from sorted list to hashmap
Map<Character, Integer> temp = new LinkedHashMap<Character, Integer>();
for (Map.Entry<Character, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static String string() {
return in.nextLine();
}
static int i() {
return in.nextInt();
}
static long l() {
return in.nextLong();
}
static int[] arrI(int N) {
int A[] = new int[N];
for (int i = 0; i < N; i++) {
A[i] = in.nextInt();
}
return A;
}
static long[] arrL(int N) {
long A[] = new long[N];
for (int i = 0; i < A.length; i++)
A[i] = in.nextLong();
return A;
}
}
class MyFastReaderB {
BufferedReader br;
StringTokenizer st;
public MyFastReaderB() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | f1573418fbb8b364293b8e5f17e864df | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.Scanner;
public class PermutationChain {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int tc= scanner.nextInt();
for (int i = 0; i<tc; i++){
int length = scanner.nextInt();
int [] array = new int[length];
System.out.println(length);
for (int j = 1; j<=length; j++){
array[j-1]=j;
System.out.print(j+ " ");
}
System.out.println();
StringBuilder first = new StringBuilder();
for (int j = 0; j<length-1; j++){
int temp = array[j];
array[j] = array[j+1];
array[j+1] = temp;
System.out.print(first);
first.append(array[j]).append(" ");
for (int t = j; t<length; t++){
System.out.print(array[t]+ " ");
}
System.out.println();
}
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | e9e335308643161299fb0e6cd8e5ea43 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
import static java.lang.Math.*;
public class Main {
static Scanner scn = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) {
int t = 1;
t = scn.nextInt();
for(int tests = 0; tests < t; tests++) solve();
out.close();
}
public static void solve() {
int n = scn.nextInt();
out.println(n);
int[] a = new int[n];
for(int i = 0; i < n; i++) a[i] = (i + 1);
printArray(a);
swap(a, 0, 1);
printArray(a);
for(int i = 3; i <= n; i++) {
swap(a, 0, i - 1);
printArray(a);
}
}
public static void printArray(int[] a) {
for(int i = 0; i < a.length; i++) out.print(a[i] + " ");
out.println();
}
public static void swap(int[] a, int i, int j) {
int t = a[i];
a[i] = a[j];
a[j] = t;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 8be720f8a767226ad9e0e654ae61b052 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
import java.util.regex.*;
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
// sorting on the basis of count of set bits
//Arrays.sort(arr, (a,b)-> Integer.compare(Integer.bitCount(b), Integer.bitCount(a)));
// provide key to pq, it will store the key according to the frequency (asc/desc)
// PriorityQueue <Integer> pq = new PriorityQueue <>((n1, n2) -> mp.get(n1)-mp.get(n2));
// Collections.sort(Arrays.asList(arr));
//char c[] = fs.next().toCharArray();
//int [] tmp = Arrays.copyOf(arr, n); // deep copy is done
// ArrayList <Integer> al = new ArrayList<>(list); // this is also a deep copy i.e changes in 'al' won't reflect in 'list'
// long ans = (long)a*(long)b; if a and b is of int type, otherwise overflow will occur
// very careful with type conversion in java.
// dont put 'long type' as index of the array,list or any Collections, only 'int' will work.
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public class Main {
static long mod = 1000000007;
static double pi = 3.14159265358979323846;
public static void main(String[] args) {
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int tc = fs.nextInt();
//int tc = 1;
for(int i = 0; i < tc; i++) {
solve(fs, out);
}
out.close();
}
public static void solve(FastScanner fs, PrintWriter out) {
int n = fs.nextInt();
int [] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = i+1;
}
out.println(n);
int k = n-2;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
out.print(arr[j] + " ");
}
out.println();
if(k < 0) {
break;
}
int tmp = arr[k];
arr[k] = arr[k+1];
arr[k+1] = tmp;
k--;
}
}
}
class Pair {
int fr;
int sc;
public Pair(int fr, int sc) {
this.fr = fr;
this.sc = sc;
}
}
class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
// read only string and breaks if finds a space
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
// read a complete line of string including spaces
String nextLine() {
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 5f17211f9b24efd25a53b4826e400d27 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
//##############################FAST READER###################################
static class FastReader{
BufferedReader br;
StringTokenizer st;
FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(st==null||!st.hasMoreElements()) {
try {
st=new StringTokenizer(br.readLine());
}catch(IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str="";
try {
str=br.readLine();
} catch(IOException e) {
e.printStackTrace();
}
return str;
}
}
//###########################################################################
static class Pair
implements Comparable<Pair>
{
int x;
int y;
Pair(int x,int y){
this.x=x;
this.y=y;
}
@Override
public int compareTo(Pair o) {
return this.x-o.x;
}
}
//##############################USEFUL FUNCTIONS#####################################
static long gcd(long a,long b) {
if(b==0) return a;
else
return gcd(b,a%b);
}
static int gcd(int a,int b) {
if(b==0) return a;
else
return gcd(b,a%b);
}
static List primeFactors(int n){
List<Integer> ar=new ArrayList<>();
int ct=0;
for(int i=2;i*i<=n;i++) {
if(n%i==0) {
ar.add(i);
ct++;
while(n%i==0) {
n/=i;
}
}
}
if(n>1) {
ct++;
ar.add(n);
}
return ar;
}
static List primeFactors(long n){
List<Integer> ar=new ArrayList<>();
int ct=0;
for(int i=2;i*i<=n;i++) {
if(n%i==0) {
ar.add(i);
ct++;
while(n%i==0) {
n/=i;
}
}
}
if(n>1) {
ct++;
ar.add((int)n);
}
return ar;
}
static long powM(long x,long y,long mod) {
if(y==0)
return 1;
long k=powM(x,y/2,mod);
if(y%2==1) {
return (((((long)1*k*k)%mod)*x)%mod);
}
else {
return (((long)1*k*k)%mod);
}
}
static int ub(ArrayList<Integer>ar,int k) {
int s=0;
int e=ar.size();
while(s!=e) {
int mid=s+e>>1;
if(ar.get(mid)<=k)
s=mid+1;
else
e=mid;
}
if(s==ar.size())
return -1;
return s;
}
static boolean isPrime(long n) {
boolean flag=true;
if(n<2)
flag=false;
else {
for(int i=2;i*i<=n;i++){
if(n%i==0) {
flag=false;
break;
}
}
}
return flag;
}
static void print(int a[]) {
for(int i=0;i<a.length;i++) {
System.out.print(a[i]+" ");
}
System.out.println();
}
static void print(long a[]) {
for(int i=0;i<a.length;i++) {
System.out.print(a[i]+" ");
}
System.out.println();
}
static void print(int a[][]) {
for(int i=0;i<a.length;i++) {
for(int j=0;j<a[0].length;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
static void print(long a[][]) {
for(int i=0;i<a.length;i++) {
for(int j=0;j<a[0].length;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
static int[] read(int[] a,FastReader sc) {
for(int i=0;i<a.length;i++) {
a[i]=sc.nextInt();
}
return a;
}
static long[] read(long[] a,FastReader sc) {
for(int i=0;i<a.length;i++) {
a[i]=sc.nextLong();
}
return a;
}
static int[][] read(int [][] a,FastReader sc){
for(int i=0;i<a.length;i++) {
for(int j=0;j<a[0].length;j++) {
a[i][j]=sc.nextInt();
}
}
return a;
}
static long[][] read(long [][] a,FastReader sc){
for(int i=0;i<a.length;i++) {
for(int j=0;j<a[0].length;j++) {
a[i][j]=sc.nextLong();
}
}
return a;
}
static void printL(List<Integer>li) {
for(int i=0;i<li.size();i++) {
System.out.print(li.get(i)+" ");
}
System.out.println();
}
static void print(List<Long>li) {
for(int i=0;i<li.size();i++) {
System.out.print(li.get(i)+" ");
}
System.out.println();
}
static String sort(String s) {
char ch[]=s.toCharArray();
Arrays.sort(ch);
return new String(ch);
}
static boolean isPalindrome(long a) {
String s=Long.toString(a);
StringBuilder rev=new StringBuilder(s);
rev.reverse();
String s1=rev.toString();
return s.equals(s1);
}
static int[] swap(int a[],int i,int j) {
int temp=a[i];
a[i]=a[j];
a[j]=temp;
return a;
}
static long[] swap(long a[],int i,int j) {
long temp=a[i];
a[i]=a[j];
a[j]=temp;
return a;
}
public static String addCharToString(String str, char c, int pos)
{
StringBuffer stringBuffer = new StringBuffer(str);
stringBuffer.insert(pos, c);
return stringBuffer.toString();
}
static int BigMod(String num, int a)
{
int res = 0;
for (int i = 0; i < num.length(); i++)
res = (res * 10 + (int)num.charAt(i));
return res;
}
static boolean isSquare(int n) {
for(int i=1;i*i<=n;i++) {
if((i*i)==n)
return true;
}
return false;
}
static long firstsetBitPower(long n)
{
int k = (int)(Math.log(n) / Math.log(2));
return 1 << k;
}
static boolean inrange(int key,int l,int r,int n) {
boolean flag=false;
if(key==0) {
if(key+1>=l&&key+1<=r) {
flag=true;
}
}
else if(key==n+1) {
if(key-1>=l&&key+1<=r) {
flag=true;
}
}
else {
if(key>=l&&key<=r) {
flag=true;
}
}
return flag;
}
static long fact(int n) {
long ans=1;
for(int i=1;i<=n;i++) {
ans*=i;
}
return ans;
}
//#############################MAIN FUNCTION##################################
public static void main(String[] args) {
try {
FastReader sc = new FastReader();
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
System.out.println(n);
for(int i=1;i<=n;i++)
System.out.print(i+" ");
System.out.println();
int tp=2;
StringBuilder sb=new StringBuilder("");
for(int i=1;i<n;i++) {
for(int j=0;j<tp;j++) {
sb.append(((j+1)%tp)+1+" ");
}
for(int j=tp;j<n;j++) {
sb.append((j+1)+" ");
}
tp++;
sb.append("\n");
}
System.out.println(sb);
}
}catch(Exception e) {return;}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 23cd0f83f0c75094f159ff995bfc1ed4 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class HackerEarth {
public static void main(String args[]){
InputReader in = new InputReader();
PrintWriter out = new PrintWriter(System.out);
Task solver = new Task();
solver.solve(1, in, out);
out.close();
}
static int mod = (int)1e9+7;
static class Task{
public void solve(int T, InputReader in, PrintWriter out) {
T = in.nextInt();
while (T-->0){
int N = in.nextInt();
out.println(N);
int a[] = new int[N];
for (int i =1; i <= N; i++){
a[i-1] = i;
}
for (int i = 0; i < N-1; i++) {
printAns(a, N, out);
int temp = a[i];
a[i] = a[i+1];
a[i+1] = temp;
}
for (int j = 0; j < N; j++){
out.print(a[j] + " ");
}
out.println();
}
}
private void printAns(int a[], int N, PrintWriter out) {
for (int j = 0; j < N; j++){
out.print(a[j] + " ");
}
out.println();
}
}
static class InputReader {
BufferedReader br;
StringTokenizer st;
public InputReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 28348988d335d443c8e6011cecf54f87 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import javax.swing.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
static AReader scan = new AReader();
static int N = 110;
static int[] a = new int[N];
static void solve() {
int n = scan.nextInt();
for(int i = 1;i<=n;i++) a[i] = i;
int k = 1;
StringBuffer sb = new StringBuffer();
sb.append(n).append("\n");
while(k<=n){
for(int i = 1;i<=n;i++) sb.append(a[i]+" ");
sb.append("\n");
int t = a[k];
a[k] = a[k+1];
a[k+1] = t;
k++;
}
System.out.print(sb.toString());
}
public static void main(String[] args) {
int T = scan.nextInt();
while (T-- > 0) {
solve();
}
}
}
class Pair{
int x,y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 215588ff7a89362f5e357d7cc8d5ebf0 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.io.PrintWriter;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int k = 0;
int[] arr = new int[n];
for (int i = 1; i <= n; i++) {
arr[i - 1] = i;
}
out.println(n);
for (int i = 0; i < n; i++) {
out.print(arr[i] + " ");
}
out.println();
for (int i = 1; i < n; i++) {
int temp = arr[i];
arr[i] = arr[i - 1];
arr[i - 1] = temp;
for (int j = 0; j < n; j++) {
out.print(arr[j] + " ");
}
out.println();
}
}
out.flush();
out.close();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 0b2ac89c319200e0b280a60b8ac0bec5 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static StringBuilder sb;
static dsu dsu;
static long fact[];
static int mod = (int) (1e9 + 7);
static void build(int node, int start, int end,int[]tree,int[]arr) {
if (start == end) {
tree[node] = arr[start];
} else {
int mid = (start + end) / 2;
int left = node * 2;
int right = node * 2 + 1;
build(left, start, mid,tree,arr);
build(right, mid + 1, end,tree,arr);
tree[node] = Math.max(tree[left], tree[right]);
}
}
static int query(int node, int start, int end, int l, int r,int[]tree,int[]arr) {
if (end < l || r < start)return Integer.MIN_VALUE;
if (start == end) {
return tree[node];
} else if (l <= start && end <= r) {
return tree[node];
} else {
int mid = (start + end) / 2;
int left = query(node * 2, start, mid, l, r,tree,arr);
int right = query(node * 2 + 1, mid + 1, end, l, r,tree,arr);
return Math.max(left, right);
}
}
static void swap(int[]arr,int i,int j){
int temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
static void solve() {
int n=i();
int[]arr=new int[n+1];
sb.append(n+"\n");
for(int i=1;i<=n;i++){
sb.append(i+" ");
arr[i]=i;
}
sb.append("\n");
for(int i=2;i<=n;i++){
swap(arr,i,i-1);
for(int j=1;j<=n;j++){
sb.append(arr[j]+" ");
}
sb.append("\n");
}
}
public static void main(String[] args) {
sb = new StringBuilder();
int test = i();
HashMap<Integer,Integer> hm=new HashMap<>();
// System.out.println(hm);
// o n e t w
fact=new long[(int)1e6+10];
fact[0]=fact[1]=1;
for(int i=2;i<fact.length;i++)
{ fact[i]=((long)(i%mod)*(long)(fact[i-1]%mod))%mod; }
int cnt=1;
while (test-- > 0) {
solve();
cnt++;
}
System.out.println(sb);
}
//**************NCR%P******************
static long ncr(int n, int r) {
if (r > n)
return (long) 0;
long res = fact[n] % mod;
// System.out.println(res);
res = ((long) (res % mod) * (long) (p(fact[r], mod - 2) % mod)) % mod;
res = ((long) (res % mod) * (long) (p(fact[n - r], mod - 2) % mod)) % mod;
// System.out.println(res);
return res;
}
static long p(long x, long y)// POWER FXN //
{
if (y == 0)
return 1;
long res = 1;
while (y > 0) {
if (y % 2 == 1) {
res = (res * x) ;
y--;
}
x = (x * x);
y = y / 2;
}
return res;
}
//**************END******************
// *************Disjoint set
// union*********//
static class dsu {
int parent[];
dsu(int n) {
parent = new int[n];
for (int i = 0; i < n; i++)
parent[i] = i;
}
int find(int a) {
if (parent[a] ==a)
return a;
else {
int x = find(parent[a]);
parent[a] = x;
return x;
}
}
void merge(int a, int b) {
a = find(a);
b = find(b);
if (a == b)
return;
parent[b] = a;
}
}
//**************PRIME FACTORIZE **********************************//
static TreeMap<Integer, Integer> prime(long n) {
TreeMap<Integer, Integer> h = new TreeMap<>();
long num = n;
for (int i = 2; i <= Math.sqrt(num); i++) {
if (n % i == 0) {
int nt = 0;
while (n % i == 0) {
n = n / i;
nt++;
}
h.put(i, nt);
}
}
if (n != 1)
h.put((int) n, 1);
return h;
}
//****CLASS PAIR ************************************************
static class Pair implements Comparable<Pair> {
int src;
long a;
long b;
Pair(int src, long a,long b) {
this.src = src;
this.a = a;
this.b = b;
}
public int compareTo(Pair o) {
return (int) (this.b - o.b);
}
}
//****CLASS PAIR **************************************************
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int Int() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String String() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return String();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
static InputReader in = new InputReader(System.in);
static OutputWriter out = new OutputWriter(System.out);
public static long[] sort(long[] a2) {
int n = a2.length;
ArrayList<Long> l = new ArrayList<>();
for (long i : a2)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a2[i] = l.get(i);
return a2;
}
public static char[] sort(char[] a2) {
int n = a2.length;
ArrayList<Character> l = new ArrayList<>();
for (char i : a2)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a2[i] = l.get(i);
return a2;
}
public static long pow(long x, long y) {
long res = 1;
while (y > 0) {
if (y % 2 != 0) {
res = (res * x);// % modulus;
y--;
}
x = (x * x);// % modulus;
y = y / 2;
}
return res;
}
//GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
public static long gcd(long x, long y) {
if (x == 0)
return y;
else
return gcd(y % x, x);
}
// ******LOWEST COMMON MULTIPLE
// *********************************************
public static long lcm(long x, long y) {
return (x * (y / gcd(x, y)));
}
//INPUT PATTERN********************************************************
public static int i() {
return in.Int();
}
public static long l() {
String s = in.String();
return Long.parseLong(s);
}
public static String s() {
return in.String();
}
public static int[] readArrayi(int n) {
int A[] = new int[n];
for (int i = 0; i < n; i++) {
A[i] = i();
}
return A;
}
public static long[] readArray(long n) {
long A[] = new long[(int) n];
for (int i = 0; i < n; i++) {
A[i] = l();
}
return A;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 90372ff567ed4e683938811e93ef48c6 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.StringTokenizer;
public class Main {
public static FastReader sc;
public static PrintWriter out;
static class FastReader{
static BufferedReader br;
static StringTokenizer str;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
static String next(){
while (str == null || !str.hasMoreElements()){
try{
str = new StringTokenizer(br.readLine());
}
catch (IOException lastMonthOfVacation){
lastMonthOfVacation.printStackTrace();
}
}
return str.nextToken();
}
static int nextInt(){
return Integer.parseInt(next());
}
static long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch (IOException lastMonthOfVacation){
lastMonthOfVacation.printStackTrace();
}
return str;
}
}
public static void main(String[] args) throws IOException{
sc = new FastReader();
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
for (int i = 0; i < t; i++){
int n = sc.nextInt();
out.println(n);
int arr[] = new int[n+1];
for (int j = 1; j< n + 1; j++){
arr[j] = j;
out.print(arr[j]+" ");
}
out.println();
for(int k = 1; k<n ; k++){
int temp = arr[k];
arr[k] = arr[k+1];
arr[k+1] = temp;
for(int p =1 ; p<=n ; p++) out.print(arr[p]+" ");
out.println();
}
}
out.close();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 4c66bb282fe6f2f9a37cf210c5e369ba | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | /*
"Everything in the universe is balanced. Every disappointment
you face in life will be balanced by something good for you!
Keep going, never give up."
Just have Patience + 1...
*/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Solution {
static int INF = (int) 1e9 + 7, MOD = 998244353;
public static void main(String[] args) throws java.lang.Exception {
out = new PrintWriter(new BufferedOutputStream(System.out));
sc = new FastReader();
int test = sc.nextInt();
for (int t = 1; t <= test; t++) {
solve(t);
}
out.close();
}
private static void solve(int t) {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = i + 1;
}
out.println(n);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
out.print(arr[j] + " ");
}
out.println();
if (i + 1 < n) {
int temp = arr[i + 1];
arr[i + 1] = arr[i];
arr[i] = temp;
}
}
}
public static FastReader sc;
public static PrintWriter out;
static class FastReader
{
BufferedReader br;
StringTokenizer str;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (str == null || !str.hasMoreElements())
{
try
{
str = new StringTokenizer(br.readLine());
}
catch (IOException lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
}
return str.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
return str;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 39c5d38c429459a6442cdac3a72ea623 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
//import java.math.BigInteger;
import java.util.*;
public class Rough {
public static void main(String[] args) throws Exception {
Scanner s = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int tc = s.nextInt();
for (int t = 1; t <= tc; t++) {
int n=s.nextInt();
int ar[]=new int[n];
for(int i=0;i<n;i++)ar[i]=i+1;
int l1=n-1;
int l2=n-2;
pw.println(n);
while(n-->0) {
for(int x:ar)pw.print(x+" ");
pw.println();
int temp=ar[l1];
ar[l1]=ar[l2];
ar[l2]=temp;
if(l1>0)l1--;
if(l2>0)l2--;
}
}
pw.close();
}
static boolean ispalin(String in) {
StringBuilder sb = new StringBuilder(in);
if (sb.reverse().toString().equals(in))
return true;
else
return false;
}
static void maping(String key, int val, HashMap<String, Integer> hm) {
if (hm.containsKey(key)) {
int valu = hm.get(key) + 1;
hm.put(key, valu);
} else
hm.put(key, val);
}
static final Random ran = new Random();
static void sort(int[] ar) {
int n = ar.length;
for (int i = 0; i < n; i++) {
int doer = ran.nextInt(n), temp = ar[doer];
ar[doer] = ar[i];
ar[i] = temp;
}
Arrays.sort(ar);
}
static void sort(long[] ar) {
int n = ar.length;
for (int i = 0; i < n; i++) {
int doer = ran.nextInt(n);
long temp = ar[doer];
ar[doer] = ar[i];
ar[i] = temp;
}
Arrays.sort(ar);
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | f8c331227124a3109a561dd42fd5d7d7 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class JavaApplication {
static BufferedReader in;
static BufferedWriter out;
static StringTokenizer st;
static String token;
String getLine() throws IOException {
return in.readLine();
}
String getToken() throws IOException {
if (st == null || !st.hasMoreTokens())
st = new StringTokenizer(getLine(), ",[] ");
return st.nextToken();
}
int getInt() throws IOException {
return Integer.parseInt(getToken());
}
long getLong() throws IOException {
return Long.parseLong(getToken());
}
char getChar() throws IOException {
if (token == null || token.length() == 0)
token = getToken();
char r = token.charAt(0);
token = token.substring(1);
return r;
}
static void print(Object s) {
try {
out.write(String.valueOf(s));
} catch (Exception ignored) {
}
}
static void println() {
try {
out.newLine();
} catch (Exception ignored) {
}
}
static void println(Object s) {
try {
out.write(String.valueOf(s));
out.newLine();
} catch (Exception ignored) {
}
}
public void Solve() throws IOException {
int t=getInt();
while(t-->0){
int n=getInt();
int[] temp=new int[n];
println(n);
for(int i=1;i<=n;i++){
print(i+" ");
temp[i-1]=i;
}
println();
int a=0;
for(int i=1;i<n;i++)
{
a=temp[i-1];
temp[i-1]=temp[i];
temp[i]=a;
for (int j=0;j<n;j++)
print(temp[j]+" ");
println();
}
}
}
public static void main(String[] args) throws java.lang.Exception {
in = new BufferedReader(new InputStreamReader(System.in));
out = new BufferedWriter(new OutputStreamWriter(System.out));
new JavaApplication().Solve();
out.flush();
return;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 44da367657528a737263615a1a17b41b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
public class PermutationChain {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
int n = Integer.parseInt(br.readLine());
bw.write(n + "\n");
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = i + 1;
}
bw.write(getString(a));
for (int i = 0; i < n - 1; i++) {
a[i] = a[i + 1];
a[i + 1] = 1;
bw.write(getString(a));
}
}
br.close();
bw.close();
}
private static String getString(int[] a) {
StringBuilder sb = new StringBuilder();
sb.append(a[0]);
for (int i = 1; i < a.length; i++) {
sb.append(" ").append(a[i]);
}
sb.append("\n");
return sb.toString();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | ba21712bac7906bb5fcd5f35c4018e27 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class Solution{
public static void main(String args[]){
Scanner in = new Scanner(System.in);
long test = in.nextLong();
while(test-- > 0){
int n = in.nextInt();
int[] ans = new int[n];
for(int i=0; i<n; i++){
ans[i]=i+1;
}
System.out.println(n);
System.out.println(Arrays.toString(ans).substring(1).replace("]", "").replace("[", "").replace(", ", " "));
for(int i=1; i<n; i++){
int temp1 = ans[i];
ans[i]=ans[i-1];
ans[i-1]=temp1;
System.out.println(Arrays.toString(ans).substring(1).replace("]", "").replace("[", "").replace(", ", " "));
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 934edd88b37f0c7605614cc0e14a4277 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int A = scn.nextInt();
StringBuilder sb = new StringBuilder("");
while(A-- > 0) {
int n = scn.nextInt();
StringBuilder str = new StringBuilder("");
int[] arr = new int[n];
for(int i=0 ; i<n ; i++)
arr[i] = i + 1;
for(int j=n-1 ; j>0 ; j--) {
for(int i=0 ; i<n ; i++) {
str.append(arr[i] + " ");
}
str.append("\n");
int temp = arr[j];
arr[j] = arr[j-1];
arr[j-1] = temp;
}
for(int i=0 ; i<n ; i++)
str.append(arr[i] + " ");
str.append("\n");
sb.append(n + "\n" + str);
}
System.out.println(sb);
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 28cf3eda48e02f5528f9480595e09ceb | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.text.*;
import java.io.*;
public class Main {
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
PrintWriter writer = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int arr[] = new int[n];
for(int i = 0; i < n; i++)arr[i] = i+1;
writer.println(n);
int point = 0;
int temp = n;
while(temp-->0) {
for(int i = 0; i < n; i++)writer.print(arr[i] + " ");
writer.println();
if(temp != 0) {
arr = swap(arr, point);
point++;
}
}
}
writer.flush();
writer.close();
}
public static boolean areEqual(ArrayList<Integer> arr1, ArrayList<Integer>arr2)
{
for(int i = 0; i < arr1.size(); i++)if(arr1.get(i) != arr2.get(i))return false;
return true;
}
private static int [] swap(int arr[], int j) {
int compl = j+1;
int ele = arr[j];
arr[j] = arr[compl];
arr[compl] = ele;
return arr;
}
private static long power (long a, long n, long p) {
long res = 1;
while(n!=0) {
if(n%2==1) {
res=(res*a)%p;
n--;
}else {
a= (a*a)%p;
n/=2;
}
}
return res;
}
private static boolean isPrime(int c) {
for (int i = 2; i*i <= c; i++) {
if(c%i==0)return false;
}
return true;
}
private static int find(int a , int arr[]) {
if(arr[a] != a) return arr[(int)a] = find(arr[(int)a], arr);
return arr[(int)a];
}
private static void union(int a, int b, int arr[]) {
int aa = find(a,arr);
int bb = find(b, arr);
arr[aa] = bb;
}
private static int gcd(int a, int b) {
if(a==0)return b;
return gcd(b%a, a);
}
public static int[] readIntArray(int size, FastReader s) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = s.nextInt();
}
return array;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
class SegmentTree{
int size;
long arr[];
SegmentTree(int n){
size = 1;
while(size<n)size*=2;
arr = new long[size*2];
}
public void build(int input[]) {
build(input, 0,0, size);
}
public void set(int i, int v) {
set(i,v,0,0,size);
}
// sum from l + r-1
public long sum(int l, int r) {
return sum(l,r,0,0,size);
}
private void build(int input[], int x, int lx, int rx) {
if(rx-lx==1) {
if(lx < input.length )
arr[x] = 1;
return;
}
int mid = (lx+rx)/2;
build(input, 2*x+1, lx, mid);
build(input,2*x+2,mid,rx);
arr[x] = arr[2*x+1] + arr[2*x+2];
}
private void set(int i, int v, int x, int lx, int rx) {
if(rx-lx==1) {
arr[x] = v;
return;
}
int mid = (lx+rx)/2;
if(i < mid) set(i, v, 2*x+1, lx, mid);
else set(i,v,2*x+2,mid,rx);
arr[x] = arr[2*x+1] + arr[2*x+2];
}
private long sum(int l, int r, int x, int lx, int rx) {
if(lx>=r || rx <= l)return 0;
if(lx>=l && rx <= r)return arr[x];
int mid = (lx+rx)/2;
long s1 = sum(l,r,2*x+1,lx,mid);
long s2 = sum(l,r,2*x+2,mid,rx);
return s2+s1;
}
// int first_above(int v, int l, int x, int lx, int rx) {
// if(arr[x].a<v)return -1;
// if(rx<=l)return -1;
// if(rx-lx==1)return lx;
// int m = (lx+rx)/2;
// int res = first_above(v,l,2*x+1,lx,m);
// if(res==-1)res = first_above(v,l,2*x+2,m,rx);
// return res;
//
// }
}
class Trie{
Trie arr[] = new Trie[26];
boolean isEnd;
Trie(){
for(int i = 0; i < 26; i++)arr[i] = null;
isEnd = false;
}
}
class Pair implements Comparable<Pair>{
long a;
long b;
long c;
Pair(long a, long b, long c){
this.a = a;
this.b = b;
this.c = c;
}
Pair(long a, long b){
this.a = a;
this.b = b;
this.c = 0;
}
@Override
public boolean equals(Object obj) {
if(this == obj) return true;
if(obj == null || obj.getClass()!= this.getClass()) return false;
Pair pair = (Pair) obj;
return (pair.a == this.a && pair.b == this.b && pair.c == this.c);
}
@Override
public int compareTo(Pair o) {
if(o.a != this.a) return Long.compare(this.a, o.a);
else
return Long.compare(this.b, o.b);
}
@Override
public String toString() {
return this.a + " " + this.b;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 9c29578190549ba300afbb7091333a5e | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) throws java.lang.Exception
{
FastReader scn = new FastReader();
int t =scn.nextInt();
while(t-- > 0){
int n = scn.nextInt();
int[] a = new int[n];
for(int i=0; i<n; i++){
a[i] = i+1;
}
System.out.println(n);
System.out.println(print(a));
for(int j=0; j<n-1; j++){
swap(a,j,j+1);
System.out.println(print(a));
}
}
}
static String print(int arr[]){
StringBuilder sb=new StringBuilder();
for(int num:arr) sb.append(num+" ");
return sb.toString();
}
static void swap(int[] a, int i,int j){
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 40de54bef8d58a00b4153116018cec93 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class B_PermutationChain {
public final static MyScanner sc = new MyScanner();
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void swap(int[] arr, int a, int b) {
int temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
}
public static void printArray(int[] arr, int a, int b) {
String ans = "";
for (int i = a; i < b; i++) {
// System.out.print(arr[i] + " ");
ans += arr[i] + " ";
}
System.out.println(ans);
}
public static void solve() {
int n = sc.nextInt();
System.out.println(n);
int[] arr = new int[n+2];
for (int i = 1; i <= n; i++) {
arr[i] = i;
}
for (int i = 1; i <= n; i++) {
printArray(arr, 1, n+1);
swap(arr, i, i+1);
}
}
public static void main(String[] args) {
int T = sc.nextInt();
for (int t = 0; t < T; t++) {
solve();
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | e04f7d17ff684a830445e7e506f6ab0b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class ProbB
{
public static void main(String[] args)
{
FastReader in = new FastReader();
int numOfTestCases = in.nextInt();
for(int w = 0; w < numOfTestCases; w++)
{
int n = in.nextInt();
System.out.println(n);
ArrayList<Integer> p = new ArrayList<Integer>();
for(int i = 1; i <= n; i++)
{
p.add(i);
}
print(p);
int c = 1;
for(int i = 0; i < n - 1; i++)
{
int temp = p.get(i + 1);
p.set(i + 1, p.get(i));
p.set(i, temp);
c++;
print(p);
}
}
}
public static void print(ArrayList<Integer> p)
{
String sp = p.toString();
sp = sp.replace(",", "");
sp = sp.replace("[", "");
sp = sp.replace("]", "");
System.out.println(sp);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] intArray(int n){
int[] arr = new int[n];
for(int i =0; i < n; i ++){
arr[i] = nextInt();
}
return arr;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | e8a4374e1140f3b5b37864800d414e3b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.lang.Math;
import java.io.*;
import java.util.Arrays;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
//StringTokenizer st = new StringTokenizer(br.readLine());
int a = Integer.parseInt(br.readLine());
for (int i = 0; i < a; i++) {
int k = Integer.parseInt(br.readLine());
pw.println(k);
int[] arr_of_nums = new int[k];
for (int j = 0; j < k; j++) {
arr_of_nums[j] = j+1;
}
pw.print(arr_of_nums[0]);
for (int j = 1; j < k; j++) {
pw.print(" " + arr_of_nums[j]);
}
pw.print("\n");
int temp = arr_of_nums[0];
arr_of_nums[0] = arr_of_nums[1];
arr_of_nums[1] = temp;
int fixedness = k - 2;
int index = 2;
pw.print(arr_of_nums[0]);
for (int j = 1; j < k; j++) {
pw.print(" " + arr_of_nums[j]);
}
pw.print("\n");
while (fixedness != 0) {
arr_of_nums[index-1] = arr_of_nums[index];
arr_of_nums[index] = temp;
fixedness -=1;
index++;
for (int num : arr_of_nums) {
pw.print(num + " ");
}
pw.print("\n");
}
}
pw.close();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 8663749fa0925e1f7f3ae2faa27b4618 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef {
public static void main(String[] args) throws java.lang.Exception {
PrintWriter out = new PrintWriter(System.out);
FastScanner sc = new FastScanner();
int tater = sc.nextInt();
while (tater-- > 0) {
int n = sc.nextInt();
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = i+1;
}
out.println(n);
int x = n;
int j = 0;
while(x-->0){
for (int i = 0 ; i < a.length ; i++){
out.print(a[i] + " ");
}
out.println();
if (x!=0){
int temp = a[j];
a[j] = a[j+1];
a[j+1] = temp;
}
j++;
}
}
out.flush();
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 692fca65cfb8941820b0d47188446e88 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static class Pair {
Pair() {
}
}
public static void main(String[] args) {
takeInput();
Scanner scn = new Scanner(System.in);
int t = 1;
t = scn.nextInt();
while (t-- > 0) {
solve(scn);
}
}
public static void solve(Scanner scn) {
int n = scn.nextInt();
System.out.println(n);
StringBuilder ans = new StringBuilder();
for (int i = n - 1; i >= 0; i--) {
int pos = i;
int num = 1;
while (pos-- > 0) {
ans.append(num + " ");
num++;
}
ans.append(n + " ");
while (num <= n - 1) {
ans.append(num + " ");
num++;
}
if (i > 0)ans.append("\n");
}
System.out.println(ans);
}
public static void takeInput() {
try {
System.setIn(new FileInputStream("input1.txt"));
System.setOut(new PrintStream(new FileOutputStream("output1.txt")));
} catch (Exception e) {
System.err.println("Error");
}
}
private static boolean isPalindrome(String s) {
int l = 0, r = s.length() - 1;
while (l < r) {
if (s.charAt(l) != s.charAt(r)) {
return false;
}
l++;
r--;
}
return true;
}
public static void swap(int x, int y) {
x = x ^ y ^ (y = x);
//x = x ^ y ^ (y = x);
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | ca7e2313e17ba5678e0785ea1ad87c7c | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class codeforces {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void help(int[] arr, int k){
int temp = arr[k];
arr[k] =arr[k-1];
arr[k-1] = temp;
}
public static void add(int[] arr,StringBuilder st ){
for(int i : arr)st.append(i).append(' ');
st.append('\n');
}
public static void main(String[] args) {
FastReader in = new FastReader();
int t =in.nextInt();
StringBuilder st = new StringBuilder();
while (t>0){
int n =in.nextInt();
int[] arr = new int[n];
st.append(n).append('\n');
for(int i=0;i<n;i++)arr[i] = i+1;
add(arr,st);
int k = n-1;
for(int i=1;i<n;i++){
help(arr,k);
k--;
add(arr,st);
}
t--;
}
System.out.println(st);
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 372730d72cad449487a9d823dc1e3f85 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class Sample {
public static void main(String[] args) {
//diptiman code
//say no to plagarism
//i dont share code nor take from anyone
Scanner xy=new Scanner(System.in);
StringBuilder ans=new StringBuilder();
//say no to plagarism
//diptiman code
//say no to plagarism
//i dont share code nor take from anyone
int tc=xy.nextInt();
while(tc>0)
{
tc--;
int n=xy.nextInt();
int A[]=new int[n];
for(int i=1;i<=n;i++)
A[i-1]=i;
int cap=0;
ans.append(n).append("\n");
while(cap<n-1)
{
for(int i=0;i<n;i++)
ans.append(A[i]).append(" ");
ans.append("\n");
int tmp=A[cap];
A[cap]=A[cap+1];
A[cap+1]=tmp;
cap++;
}
for(int i=0;i<n;i++)
ans.append(A[i]).append(" ");
ans.append("\n");
}
System.out.print(ans);
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 7db53620d9af8294be9985b01e20f3ef | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Reader r = new Reader();
int t = r.tN();
while(t-->0){
//solve(r);
//solveB(r);
solveC(r);
//solveD(r);
}
}
public static void solve(Reader r) throws Exception{
int n = r.tN();
// int[] arr = r.tIA();
if(n==1){
System.out.println(2);
return;
}
if(n==2){
System.out.println(1);
return;
}
int ans = n/3;
if(n%3!=0) ans++;
System.out.println(ans);
}
public static void solveB(Reader r) throws Exception{
int n = r.tN();
int[] arr = r.tIA();
}
public static void solveC(Reader r) throws Exception{
int n = r.tN();
//int[] arr = r.tIA();
System.out.println(n);
for(int i = 1; i<=n; i++){
int count = 2;
StringBuilder sb = new StringBuilder();
for(int j = 1; j<=n; j++){
if(j==i){
sb.append(1+" ");
} else{
sb.append(count+" ");
count++;
}
}
System.out.println(sb);
}
}
public static void solveD(Reader r) throws Exception{
int n= r.tN();
}
public static class Graph{
ArrayList<Integer>[] graph;
int nodeCount;
int edgesCount;
int[] nodeValue;
//boolean[] visited = new boolean[n+1];
public Graph(int n){
nodeCount = n;
graph = new ArrayList[n+1];
initGraph();
}
public void setNodeValue(int[] arr){
nodeValue = new int[nodeCount+1];
for(int i = 1; i<=nodeCount; i++){
nodeValue[i] = arr[i-1];
}
}
private void initGraph(){
for(int i = 0; i<=nodeCount; i++){
graph[i] = new ArrayList<>();
}
}
public void addOrientedEdge(int u, int v){
graph[u].add(v);
}
public void addEdge(int u, int v){
graph[u].add(v);
graph[v].add(u);
}
}
public static class myArray{
public static void swap(int[] arr, int i, int j){
int temp = arr[i] ;
arr[i] = arr[j];
arr[j] = temp;
}
public static void reverseArr(int[] arr){
int i = 0;
int j = arr.length-1;
while(i<j){
swap(arr,i,j);
i++;
j--;
}
}
}
public static class SegmentTree{
int[] tree;
int count;
public SegmentTree(int[] arr){
count = arr.length;
tree = new int[4*count];
build(arr,0,count-1,1);
}
public SegmentTree(int n){
count = n;
int[] arr = new int[n];
tree = new int[n*4];
build(arr,0,count-1,1);
}
public void build(int[] arr, int s, int e, int node){
if(s==e){
tree[node] = arr[s];
return;
}
int mid = s+(e-s)/2;
build(arr,s,mid,node*2);
build(arr,mid+1,e,node*2+1);
tree[node]= tree[node*2]+tree[node*2+1];
}
void update(int pos ,int val){
update(1,0,count-1,pos,val);
}
private void update(int node, int s , int e , int pos, int val){
if(e<pos||s>pos) return;
if(s==e && e==pos){
// System.out.println("Updating "+ s + " to " + e + " : " + val);
tree[node] = val;
return;
}
int mid = s+(e-s)/2;
update(node*2,s,mid,pos,val);
update(node*2+1,mid+1,e,pos,val);
tree[node]= tree[node*2]+tree[node*2+1];
//System.out.println("Updating "+ s + " to " + e + " : " + tree[node]);
}
int get(int l , int r){
return get(1,0,count-1,l,r);
}
private int get(int node , int s , int e, int l , int r){
if(e<l||s>r) return 0;
if(s==e ||(s>=l && e<=r)){
//System.out.println("returning sum from " + s + " to " +e " : " + tree[node]);
return tree[node];
}
int mid = s + (e-s)/2;
int left = get(node*2,s,mid,l,r);
int right = get(node*2+1,mid+1,e,l,r);
return left + right;
}
}
public static class Pair{
long val;
int sval;
public Pair(long v , int i){
this.val = v;
this.sval = i;
}
}
public static class Reader {
BufferedReader br;
public Reader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public int tN() throws Exception {
return Integer.parseInt(takeString());
}
public long tL() throws Exception {
return Long.parseLong(takeString());
}
public int[] tIA() throws Exception {
String[] arr = tSA();
int[] vals = convertStringIntoArray(arr);
return vals;
}
public long[] tLA() throws Exception{
String[] arr = br.readLine().split(" ");
long[] res = new long[arr.length];
for(int i = 0; i<arr.length; i++){
res[i] = Long.parseLong(arr[i]);
}
return res;
}
public String takeString() throws Exception{
return br.readLine();
}
public String[] tSA() throws Exception{
return takeString().split(" ");
}
public static int[] convertStringIntoArray(String[] nums){
int[] arr = new int[nums.length];
for(int i =0; i<nums.length; i++){
arr[i] = Integer.parseInt(nums[i]);
}
return arr;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | abcd1f785b2321c5c2108e9f267e6054 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class d3
{
static class Node{
int left;
int right;
public Node(int left,int right){
this.left = left;
this.right = right;
}
}
static class pair1{
int number;
int nothappy;
int size ;
pair1(int a,int b,int c){
this.number = a;
this.nothappy = b;
this.size = c;
}
}
public static void main(String[] args) throws Exception
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
while(t-->0){
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int[] arr = new int[n];
bw.write(n+"\n");
for(int i = 0;i<n;i++){
arr[i] = i+1;
bw.write(arr[i]+" ");
}
bw.write("\n");
arr[1] = 1;
arr[0] = 2;
for(int i = 1;i<n-1;i++){
for(int j = 0;j<n;j++){
bw.write(arr[j]+" ");
}
bw.write("\n");
int temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
}
// int temp = arr[n-2];
// arr[n-2] = arr[n-1];
// arr[n-1] = temp;
for(int j = 0;j<n;j++){
bw.write(arr[j]+" ");
}
bw.write("\n");
}
bw.flush();
}
public static void fill(int x,int l,int r,long[] arr,long[] seg){
if(l == r){
seg[x] = arr[l]; //leaf element
return;
}
int mid = (l+r)/2;
fill(2*x,l,mid,arr,seg);
fill(2*x+1,mid+1,r,arr,seg);
seg[x] =seg[2*x]+seg[2*x+1];
}
public static long sum(int x,int l,int r,int lr,int rr,long[] arr,long[] seg){
//completely outside
if(r < lr || rr < l){
return 0;
}
// lr.....l....r......rr
if(lr <= l && r <= rr){
return seg[x];
}
int mid = (l+r)/2;
long a = 0;
a += sum(2*x,l,mid,lr,rr,arr,seg);
a += sum(2*x+1,mid+1,r,lr,rr,arr,seg);
return a;
}
public static void update(int x,int l,int r,int i,int val,long[] arr,long seg[]){
if(l == r){
seg[x] = val;
return;
}
int mid = (l+r)/2;
if(l <= i && i <= mid){
update(2*x,l,mid,i,val,arr,seg);
}
else{
update(2*x+1,mid+1,r,i,val,arr,seg);
}
seg[x] = (seg[2*x]+seg[2*x+1]);
}
// public static void mergesort(int[] arr,int low,int high){
// if(low == high){
// return;
// }
// int mid = (low+high)/2;
// mergesort(arr, low, mid);
// mergesort(arr, mid+1, high);
// merge(arr,low,high);
// }
// public static void merge(int[] arr,int low,int high){
// int mid = (low+high)/2;
// int p1 = low;
// int p2 = mid+1;
// int k = 0;
// int[] temp = new int[high-low+1];
// while(p1 <= mid && p2 <= high){
// if(arr[p1] <= arr[p2]){
// temp[k++] = arr[p1++];
// }
// else{
// temp[k++] = arr[p2++];
// count += (mid-p1+1);
// }
// }
// while(p1 <= mid){
// temp[k++] = arr[p1++];
// }
// while(p2 <= high){
// temp[k++] = arr[p2++];
// }
// for(int i = 0;i<temp.length;i++){
// arr[low+i] = temp[i];
// }
// }
public static boolean check2(long mid,long n,long r,long c){
long noofrows = mid/r;
long percolumn = mid/c;
long cur = noofrows*percolumn;
if(cur >= n){
return true;
}
return false;
}
public static int ceil1(int[] arr,int low,int high,int ele){
while(low <= high){
int mid = (low+high)/2;
if(arr[mid] < ele){
low = mid+1;
}
else{
high = mid-1;
}
}
return low;
}
public static int floor1(int[] arr,int low,int high,int ele){
while(low <= high){
int mid = (low+high)/2;
if(arr[mid] <= ele){
low = mid+1;
}
else{
high = mid-1;
}
}
return low;
}
public static int dfs(int[] arr,int i,int[] ans,boolean[] visited){
if(i >= arr.length){
return 0;
}
if(visited[i]){
return ans[i];
}
visited[i] = true;
ans[i] += arr[i]+dfs(arr,i+arr[i],ans,visited);
return ans[i];
}
// public static int dfs(ArrayList<ArrayList<Integer>> graph,int i,boolean[] visited){
// if(!visited[i]){
// visited[i]=true;
// for(int child : graph.get(i)){
// if(!visited[child]){
// return dfs(graph, child, visited);
// }
// else{
// return child;
// }
// }
// }
// else{
// return i;
// }
// return 0;
// }
public static boolean recur(int i,int[] arr){
if(i < 0){
return true;
}
if(arr[i] != arr[arr.length-1-i]){
return false;
}
if(recur(i-1,arr)){
return true;
}else{
return false;
}
}
public static long power(long a,long b,long mod){
long res = 1L;
while(b > 0){
if(b%2 == 1){
res = (res%mod * a%mod)%mod;
}
b /= 2;
a = (a*a)%mod;
}
return res;
}
public static boolean comp(String s1,String s2){
int count = 0;
for(int i = 0;i<2;i++){
if(s1.charAt(i) == s2.charAt(1-i)){
count++;
}
else{
return false;
}
}
// System.out.println(s1+" "+s2);
return true;
}
public static boolean palin1(String s){
int n = s.length()-1;
int mid = (n+1)/2;
for(int i = 0;i<mid;i++){
if(s.charAt(i) != s.charAt(n-i)){
return false;
}
}
return true;
}
static class pair{
int a;
int b;
pair(int a,int b){
this.a = a;
this.b = b;
}
}
public static boolean checksum(int n){
int temp = 0;
while(n > 0){
temp += n%10;
n /= 10;
}
return (temp == 10)?true:false;
}
public static int ceil(ArrayList<Long> al,long val){
int low = 0;
int high = al.size()-1;
int pos = 0;
while(low <= high){
int mid = (low+high)/2;
if(al.get(mid) == val){
pos = Math.max(pos,mid);
low = mid+1;
}
else if(al.get(mid) < val){
low = mid+1;
}
else{
high = mid-1;
}
}
return pos;
}
public static int floor(ArrayList<Long> al,long val){
int low = 0;
int high = al.size()-1;
int pos = high;
while(low <= high){
int mid = (low+high)/2;
if(al.get(mid) == val){
pos = Math.min(pos,mid);
high = mid-1;
}
else if(al.get(mid) < val){
low = mid+1;
}
else{
high = mid-1;
}
}
return pos;
}
public static boolean palin(String s){
for(int i = 0;i<s.length()/2;i++){
if(s.charAt(i) == s.charAt(s.length()-i-1)){
continue;
}
else{
return false;
}
}
return true;
}
public static long gcd(long a,long b){
if(b == 0L){
return a;
}
return gcd(b,a%b);
}
public static long fin(long n){
return ((long)n)*(n+1)/2;
}
public static int bceil(ArrayList<Integer> al,int val,int low,int high){
int index = 0;
while(low <= high){
int mid = low + (high-low)/2;
if(al.get(mid) <= val){
index = Math.max(mid,index);
low = mid+1;
}
else{
high = mid-1;
}
}
return index;
}
public static int bfloor(ArrayList<Integer> al,int val){
int low = 0;
int high = al.size()-1;
int index = -1;
while(low <= high){
int mid = low + (high-low)/2;
int mval = al.get(mid);
if(mval >= val){
high = mid-1;
index = Math.max(index,mid);
}
else{
low = mid+1;
}
}
return index;
}
public static void primefact(int n){
for(int i = 2;i*i<=n;i++){
if(n%i == 0){
while(n%i == 0){
n /= i;
System.out.println(i);
}
}
}
System.out.println(n);
}
public static boolean prime(long n){
int count = 0;
for(long i = 1;i*i <= n;i++){
if(n%i == 0){
if(i*i == n){
count++;
}
else{
count += 2;
}
}
}
if(count == 2){
return true;
}
return false;
}
public static long fact1(int n){
long ans = 1L;
for(int i = 2;i<=n;i++){
ans *= i;
}
return ans;
}
public static String reverse(String s){
StringBuilder sb = new StringBuilder(s);
return sb.reverse().toString();
}
public static int find(int n){
return n*(n+1)/2;
}
public static double calc(int x1,int x2,int y1,int y2){
return Math.sqrt(Math.pow(x1-x2,2)+Math.pow(y1-y2,2));
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 89049776b03f3ed4f1cbf0b3c71ee5e2 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.List;
import java.util.Objects;
import java.util.StringTokenizer;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
public class B_PermutationsSeq {
public static void main(String[] args) throws IOException {
BufferedScanner input = new BufferedScanner();
BufferedOutput output = new BufferedOutput();
int tasks = input.nextInt();
while (tasks-- > 0) {
int size = input.nextInt();
List<Integer> permutation = IntStream.range(1, size + 1).boxed().collect(Collectors.toList());
output.println(size);
output.println(permutation.stream().map(Objects::toString).collect(Collectors.joining(" ")));
for (int i = size - 1; i > 0; i--) {
permutation.set(i, permutation.get(0));
permutation.set(0, i + 1);
output.println(permutation.stream().map(Objects::toString).collect(Collectors.joining(" ")));
}
}
output.flush();
}
private static class BufferedScanner {
private final BufferedReader reader;
private StringTokenizer tokenizer;
public BufferedScanner() {
reader = new BufferedReader(new InputStreamReader(System.in));
}
public int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
public String nextLine() throws IOException {
tokenizer = null;
return reader.readLine();
}
private String nextToken() throws IOException {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
}
private static class BufferedOutput {
private final StringBuilder result;
public BufferedOutput() {
result = new StringBuilder();
}
public void print(Object object) {
result.append(object);
}
public void println(Object object) {
result.append(object).append("\n");
}
public void flush() {
System.out.println(result);
System.out.flush();
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 20b038fd6970105dcd1fea4e06f584f6 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | //<———My cp————
//https://takeuforward.org/interview-experience/strivers-cp-sheet/?utm_source=youtube&utm_medium=striver&utm_campaign=yt_video
import java.util.*;
import java.io.*;
public class B_Permutation_Chain{
static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) throws Exception{
FastReader fr = new FastReader(System.in);
int t = fr.nextInt();
while(t-->0){
int n = fr.nextInt();
int[] vals = new int[n];
for(int i = 0;i<vals.length;i++){
vals[i]=i+1;
}
pw.println(n);
for(int i = 0;i<n;i++){
pw.print(vals[i]+" ");
}
pw.println("");
for(int i =n-1;i>0;i--){
int temp = vals[i-1];
vals[i-1]=vals[i];
vals[i]=temp;
for(int k = 0;k<n;k++){
pw.print(vals[k]+" ");
}
pw.println("");
}
}
pw.close();;
}
public static int[] sort(int[] vals){
ArrayList<Integer> values = new ArrayList<>();
for(int i = 0;i<vals.length;i++){
values.add(vals[i]);
}
Collections.sort(values);
for(int i =0;i<values.size();i++){
vals[i] = values.get(i);
}
return vals;
}
public static long[] sort(long[] vals){
ArrayList<Long> values = new ArrayList<>();
for(int i = 0;i<vals.length;i++){
values.add(vals[i]);
}
Collections.sort(values);
for(int i =0;i<values.size();i++){
vals[i] = values.get(i);
}
return vals;
}
public static void reverseArray(long[] vals){
int startIndex = 0;
int endIndex = vals.length-1;
while(startIndex<=endIndex){
long temp = vals[startIndex];
vals[startIndex] = vals[endIndex];
vals[endIndex] = temp;
startIndex++;
endIndex--;
}
}
public static void reverseArray(int[] vals){
int startIndex = 0;
int endIndex = vals.length-1;
while(startIndex<=endIndex){
int temp = vals[startIndex];
vals[startIndex] = vals[endIndex];
vals[endIndex] = temp;
startIndex++;
endIndex--;
}
}
static class FastReader{
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
public static int GCD(int numA, int numB){
if(numA==0){
return numB;
}else if(numB==0){
return numA;
}else{
if(numA>numB){
return GCD(numA%numB,numB);
}else{
return GCD(numA,numB%numA);
}
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 1f947186c855cc48ae989cbf78179816 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.Arrays;
public class Main {
public static void main(String[] args) throws IOException {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int tests = in.nextInt();
for (int i = 0;i < tests;i++) {
solve (in, out);
}
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
// main code ?
static void solve (FastScanner in, PrintWriter out) {
int n = in.nextInt(), arr[] = new int[n];
out.println(n);
for (int i = 1;i <= n;i++) arr[i-1] = i;
for (int i = 0;i < n;i++) {
for (int j = 0;j < n;j++) {
out.print(arr[j] + " ");
}
if (i < n-1) {
int t = arr[i+1];
arr[i+1] = arr[i];
arr[i] = t;
}
out.println();
}
out.flush();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 9ec623f10fa282a3e6c214a76691f5fb | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | // Author - Diwakar Singh
//----------------------------------------------------------------------Libraries----------------------------------------------------------------------
import java.io.*;
import java.util.*;
import java.math.*;
//--------------------------------------------------------------------fast functions----------------------------------------------------------------------
// fastPower(int a, int n) --> Power of a number
// cnt_div(int n) --> count divisors
// GcdOfArray(array, 0) ---> get gcd of whole array
//----------------------------------------------------------------------Main Class--------------------------------------------------------------------------
public class Main
{
static PrintWriter out=new PrintWriter((System.out));
static int mod = 1000000007, imax=Integer.MAX_VALUE, imin=Integer.MIN_VALUE, e9=1000000009;
static long lmax=Long.MAX_VALUE, lmin=Long.MIN_VALUE;
public static int solver(long n){
int max = imin;
for (int i=1; i<=Math.sqrt(n); i++)
{
if(n%i==0){
max = Math.max(max, i);
}
}
return max;
}
//-------------------------------------------------------------------Driver code------------------------------------------------------------------------------
public static void main(String args[])throws IOException
{
Reader sc=new Reader();
int t=sc.i();
while(t-->0){
int n = sc.i(), start=0, end=n-1;
int arr[] = new int[n];
for(int i=0;i<n;i++){
arr[i] = i+1;
}
out.println(n);
for(int i=0;i<arr.length;i++){
out.print(arr[i]+" ");
}
out.println();
while(start<end){
int temp = arr[start];
arr[start] =arr[end];
arr[end] = temp;
for(int i=0;i<arr.length;i++){
out.print(arr[i]+" ");
}
start++;
out.println();
}
}
out.close();
}
// -----------------------------------------------------------------fast functions---------------------------------------------------------------------
// static int __gcd(int a, int b)
// {
// return b == 0? a:__gcd(b, a % b);
// }
// static int GcdOfArray(int[] arr, int idx)
// {
// if (idx == arr.length - 1) {
// return arr[idx];
// }
// int a = arr[idx];
// int b = GcdOfArray(arr, idx + 1);
// return __gcd(a, b); // __gcd(a,b) is inbuilt library function
// }
// static int fastPower(int a, int n){
// if(n==0) return 1;
// if(n==1) return a;
// if((n&1)==1) return a*fastPower(a, n/2)*fastPower(a, n/2);
// return fastPower(a,n/2)*fastPower(a, n/2);
// }
static boolean prime(int n){
int count=0;
for(int i=1;i*i<=n;i++){
if(n%i==0){
count++;
if(i!=n/i) count++;
}
if(count>2) return false;
}
return true;
}
//---------------------------------------------------------------------Reader for input--------------------------------------------------------------------
static class Reader
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
public String s()
{
while(!st.hasMoreTokens())
{
try
{
st=new StringTokenizer(br.readLine());
}
catch(Exception e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
public int i()
{
return Integer.parseInt(s());
}
public long l()
{
return Long.parseLong(s());
}
public double d()
{
return Double.parseDouble(s());
}
public String ns()
{
try
{
return br.readLine();
}
catch(Exception e)
{
e.printStackTrace();
}
return null;
}
public boolean b()
{
String next=null;
try
{
next=br.readLine();
}
catch(Exception e)
{
}
if(next==null)
{
return false;
}
st=new StringTokenizer(next);
return true;
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 0438b3580a8a00ba459dc02a07fc5c74 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static long mod = (int)1e9+7;
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
// int t=1;
while(t-->0)
{
int n = sc.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++)
{
arr[i] = i + 1;
}
out.println(n);
for(int x:arr)
{
out.print(x+" ");
}
out.println();
for(int i=0;i<n-1;i++)
{
int temp = arr[i];
arr[i] = arr[n - 1];
arr[n - 1] = temp;
for(int x:arr)
{
out.print(x+" ");
}
out.println();
}
}
out.flush();
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
static class FenwickTree
{
//Binary Indexed Tree
//1 indexed
public int[] tree;
public int size;
public FenwickTree(int size)
{
this.size = size;
tree = new int[size+5];
}
public void add(int i, int v)
{
while(i <= size)
{
tree[i] += v;
i += i&-i;
}
}
public int find(int i)
{
int res = 0;
while(i >= 1)
{
res += tree[i];
i -= i&-i;
}
return res;
}
public int find(int l, int r)
{
return find(r)-find(l-1);
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
private static long mergeAndCount(int[] arr, int l,
int m, int r)
{
// Left subarray
int[] left = Arrays.copyOfRange(arr, l, m + 1);
// Right subarray
int[] right = Arrays.copyOfRange(arr, m + 1, r + 1);
int i = 0, j = 0, k = l;long swaps = 0;
while (i < left.length && j < right.length) {
if (left[i] < right[j])
arr[k++] = left[i++];
else {
arr[k++] = right[j++];
swaps += (m + 1) - (l + i);
}
}
while (i < left.length)
arr[k++] = left[i++];
while (j < right.length)
arr[k++] = right[j++];
return swaps;
}
// Merge sort function
private static long mergeSortAndCount(int[] arr, int l,
int r)
{
// Keeps track of the inversion count at a
// particular node of the recursion tree
long count = 0;
if (l < r) {
int m = (l + r) / 2;
// Total inversion count = left subarray count
// + right subarray count + merge count
// Left subarray count
count += mergeSortAndCount(arr, l, m);
// Right subarray count
count += mergeSortAndCount(arr, m + 1, r);
// Merge count
count += mergeAndCount(arr, l, m, r);
}
return count;
}
static void my_sort(long[] arr)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list);
for(int i=0;i<arr.length;i++)
{
arr[i] = list.get(i);
}
}
static void reverse_sorted(int[] arr)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list , Collections.reverseOrder());
for(int i=0;i<arr.length;i++)
{
arr[i] = list.get(i);
}
}
static int LowerBound(int a[], int x) { // x is the target value or key
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
static int UpperBound(ArrayList<Integer> list, int x) {// x is the key or target value
int l=-1,r=list.size();
while(l+1<r) {
int m=(l+r)>>>1;
if(list.get(m)<=x) l=m;
else r=m;
}
return l+1;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static class Queue_Pair implements Comparable<Queue_Pair> {
int first , second;
public Queue_Pair(int first, int second) {
this.first=first;
this.second=second;
}
public int compareTo(Queue_Pair o) {
return Integer.compare(o.first, first);
}
}
static void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
static void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
static boolean isPalindrome(String str)
{
// Pointers pointing to the beginning
// and the end of the string
int i = 0, j = str.length() - 1;
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str.charAt(i) != str.charAt(j))
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;
}
static boolean palindrome_array(char arr[], int n)
{
// Initialise flag to zero.
int flag = 0;
// Loop till array size n/2.
for (int i = 0; i <= n / 2 && n != 0; i++) {
// Check if first and last element are different
// Then set flag to 1.
if (arr[i] != arr[n - i - 1]) {
flag = 1;
break;
}
}
// If flag is set then print Not Palindrome
// else print Palindrome.
if (flag == 1)
return false;
else
return true;
}
static boolean allElementsEqual(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
static boolean allElementsDistinct(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]!=arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
public static void reverse_Long(long[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
long temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
static boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
static boolean isReverseSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] < a[i + 1]) {
return false;
}
}
return true;
}
static int[] rearrangeEvenAndOdd(int arr[], int n)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
int[] array = list.stream().mapToInt(i->i).toArray();
return array;
}
static long[] rearrangeEvenAndOddLong(long arr[], int n)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
long[] array = list.stream().mapToLong(i->i).toArray();
return array;
}
static boolean isPrime(long n)
{
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (long i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static long getSum(long n)
{
long sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n/10;
}
return sum;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcdLong(long a, long b)
{
if (b == 0)
return a;
return gcdLong(b, a % b);
}
static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static int countDigit(int n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
}
class Pair
{
int first;
int second;
Pair(int first , int second)
{
this.first = first;
this.second = second;
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | fe9cc9c7ac3a7386506fa3f8b736126d | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static int mod = 998244353;
//static int mod = (int)1e9+7;
static boolean[] prime = new boolean[1];
static int[][] dir1 = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
static int[][] dir2 = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
static int inf = 0x3f3f3f3f;
static {
for (int i = 2; i < prime.length; i++)
prime[i] = true;
for (int i = 2; i < prime.length; i++) {
if (prime[i]) {
for (int k = 2; i * k < prime.length; k++) {
prime[i * k] = false;
}
}
}
}
static class DSU {
int[] fa;
DSU(int n) {
fa = new int[n];
for (int i = 0; i < n; i++)
fa[i] = i;
}
int find(int t) {
if (t != fa[t])
fa[t] = find(fa[t]);
return fa[t];
}
void join(int x, int y) {
x = find(x);
y = find(y);
fa[x] = y;
}
}
static List<Integer>[] lists;
static void init(int n) {
lists = new List[n];
for(int i = 0; i< n;i++){
lists[i] = new ArrayList<>();
}
}
static class LCA{
int[] dep;
int[][] fa;
int[] log;
boolean[] v;
public LCA(int n){
dep = new int[n+5];
log = new int[n+5];
fa = new int[n+5][31];
v = new boolean[n+5];
for (int i = 2; i <= n; ++i) {
log[i] = log[i/2] + 1;
}
dfs(1,0);
}
private void dfs(int cur, int pre){
if(v[cur]) return;
v[cur] = true;
dep[cur] = dep[pre]+1;
fa[cur][0] = pre;
for (int i = 1; i <= log[dep[cur]]; ++i) {
fa[cur][i] = fa[fa[cur][i - 1]][i - 1];
}
for(int i : lists[cur]){
dfs(i,cur);
}
}
private int lca(int a, int b){
if(dep[a] > dep[b]){
int t = a;
a = b;
b = t;
}
while (dep[a] != dep[b]){
b = fa[b][log[dep[b] - dep[a]]];
}
if(a == b) return a;
for (int k = log[dep[a]]; k >= 0; k--) {
if (fa[a][k] != fa[b][k]) {
a = fa[a][k]; b = fa[b][k];
}
}
return fa[a][0];
}
}
static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
static int get() throws Exception {
String ss = bf.readLine();
if (ss.contains(" "))
ss = ss.substring(0, ss.indexOf(" "));
return Integer.parseInt(ss);
}
static long getx() throws Exception {
String ss = bf.readLine();
if (ss.contains(" "))
ss = ss.substring(0, ss.indexOf(" "));
return Long.parseLong(ss);
}
static int[] getint() throws Exception {
String[] s = bf.readLine().split(" ");
int[] a = new int[s.length];
for (int i = 0; i < a.length; i++) {
a[i] = Integer.parseInt(s[i]);
}
return a;
}
static long[] getlong() throws Exception {
String[] s = bf.readLine().split(" ");
long[] a = new long[s.length];
for (int i = 0; i < a.length; i++) {
a[i] = Long.parseLong(s[i]);
}
return a;
}
static String getstr() throws Exception {
return bf.readLine();
}
static void println() throws Exception {
bw.write("\n");
}
static void print(int a) throws Exception {
bw.write(a + "\n");
}
static void print(long a) throws Exception {
bw.write(a + "\n");
}
static void print(String a) throws Exception {
bw.write(a + "\n");
}
static void print(int[] a) throws Exception {
for (int i : a) {
bw.write(i + " ");
}
println();
}
static void print(long[] a) throws Exception {
for (long i : a) {
bw.write(i + " ");
}
println();
}
static void print(int[][] a) throws Exception {
for (int i[] : a)
print(i);
}
static void print(long[][] a) throws Exception {
for (long[] i : a)
print(i);
}
static void print(char[] a) throws Exception {
for (char i : a) {
bw.write(i + "");
}
println();
}
static long pow(long a, long b) {
long ans = 1;
while (b > 0) {
if ((b & 1) == 1) {
ans *= a;
}
a *= a;
b >>= 1;
}
return ans;
}
static int powmod(long a, long b, int mod) {
long ans = 1;
while (b > 0) {
if ((b & 1) == 1) {
ans = ans * a % mod;
}
a = a * a % mod;
b >>= 1;
}
return (int) ans;
}
static void sort(int[] a) {
int n = a.length;
Integer[] b = new Integer[n];
for (int i = 0; i < n; i++)
b[i] = a[i];
Arrays.sort(b);
for (int i = 0; i < n; i++)
a[i] = b[i];
}
static void sort(long[] a) {
int n = a.length;
Long[] b = new Long[n];
for (int i = 0; i < n; i++)
b[i] = a[i];
Arrays.sort(b);
for (int i = 0; i < n; i++)
a[i] = b[i];
}
static void resort(int[] a) {
int n = a.length;
Integer[] b = new Integer[n];
for (int i = 0; i < n; i++)
b[i] = a[i];
Arrays.sort(b);
for (int i = 0; i < n; i++)
a[i] = b[n - 1 - i];
}
static void resort(long[] a) {
int n = a.length;
Long[] b = new Long[n];
for (int i = 0; i < n; i++)
b[i] = a[i];
Arrays.sort(b);
for (int i = 0; i < n; i++)
a[i] = b[n - 1 - i];
}
static int max(int a, int b) {
return Math.max(a, b);
}
static int min(int a, int b) {
return Math.min(a, b);
}
static long max(long a, long b) {
return Math.max(a, b);
}
static long min(long a, long b) {
return Math.min(a, b);
}
static int max(int[] a) {
int max = a[0];
for (int i : a)
max = max(max, i);
return max;
}
static int min(int[] a) {
int min = a[0];
for (int i : a)
min = min(min, i);
return min;
}
static long max(long[] a) {
long max = a[0];
for (long i : a)
max = max(max, i);
return max;
}
static long min(long[] a) {
long min = a[0];
for (long i : a)
min = min(min, i);
return min;
}
static int abs(int a) {
return Math.abs(a);
}
static long abs(long a) {
return Math.abs(a);
}
static void yes() throws Exception {
print("Yes");
}
static void no() throws Exception {
print("No");
}
static int[] getarr(List<Integer> list) {
int n = list.size();
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = list.get(i);
return a;
}
static int[] ans;
public static void main(String[] args) throws Exception {
int T = 1;
T = get();
while (T-- > 0) {
int n = get();
List<int[]> list = new ArrayList<>();
int[] a = new int[n];
for(int i = 0;i < n;i++){
a[i] = i+1;
}
list.add(a);
int k = 0,f = 0;
while (k < n-1) {
int[] c = list.get(list.size() - 1).clone();
swap(c, k, k + 1);
k++;
list.add(c);
}
print(list.size());
for(int i[] : list) print(i);
}
bw.flush();
}
static void swap(int[] a, int i, int j){
int t = a[i];
a[i] = a[j];
a[j] = t;
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | ed38a268a3de0d371decbbd95ad1d223 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
import java.lang.*;
public class Main {
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
public static int cnt;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String args[]) {
int mod = 1000000007;
MyScanner sc = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n+1];
out.println(n);
for(int i=1;i<=n;i++){
out.print(i+" ");
a[i] = i;
}
out.println();
for(int i=1;i<n;i++){
int t1 = a[i];
a[i] = a[i+1];
a[i+1] = t1;
for(int j=1;j<=n;j++){
out.print(a[j]+" ");
}
out.println();
}
}
out.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 0e5cc7e4104b9b6c5a39fdbddda30749 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.lang.*;
public class Solution{
static int mod=(int)1e9+7;
static int mod1=998244353;
static FastScanner sc = new FastScanner();
static StringBuffer as=new StringBuffer("");
public static void solve()
{
int n=sc.nextInt();
int[] ans=new int[n];
// System.out.println(n);
as.append(n+"\n");
for (int i=0;i<n;i++ ) {
ans[i]=i+1;
}
for (int i=0;i<n;i++) {
int temp=ans[i];
ans[i]=ans[0];
ans[0]=temp;
for (int j=0;j<n;j++ ) {
// System.out.print(ans[j]+" ");
as.append(ans[j]+" ");
}
// System.out.println("");
as.append("\n");
}
}
public static void main(String[] args) {
int t = sc.nextInt();
// int t=1;
outer: for (int tt = 0; tt < t; tt++) {
solve();
}
System.out.println(as);
}
static int[] leftRotate(int arr[], int n, int k)
{
int[] b=new int[n];
int mod = k % n;
for (int i = 0; i < n; ++i)
b[i]=arr[(i + mod) % n] ;
return b;
}
static boolean isSubSequence(String str1, String str2,
int m, int n)
{
int j = 0;
for (int i = 0; i < n && j < m; i++)
if (str1.charAt(j) == str2.charAt(i))
j++;
return (j == m);
}
static int reverseDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
/* Function to check if n is Palindrome*/
static boolean isPalindrome(int n)
{
// get the reverse of n
int rev_n = reverseDigits(n);
// Check if rev_n and n are same or not.
if (rev_n == n)
return true;
else
return false;
}
static boolean isPrime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static int[] LPS(String s){
int[] lps=new int[s.length()];
int i=0,j=1;
while (j<s.length()) {
if(s.charAt(i)==s.charAt(j)){
lps[j]=i+1;
i++;
j++;
continue;
}else{
if (i==0) {
j++;
continue;
}
i=lps[i-1];
while(s.charAt(i)!=s.charAt(j) && i!=0) {
i=lps[i-1];
}
if(s.charAt(i)==s.charAt(j)){
lps[j]=i+1;
i++;
}
j++;
}
}
return lps;
}
static long getPairsCount(int n, double sum,int[] arr)
{
HashMap<Double, Integer> hm = new HashMap<>();
for (int i = 0; i < n; i++) {
if (!hm.containsKey((double)arr[i]))
hm.put((double)arr[i], 0);
hm.put((double)arr[i], hm.get((double)arr[i]) + 1);
}
long twice_count = 0;
for (int i = 0; i < n; i++) {
if (hm.get(sum - arr[i]) != null)
twice_count += hm.get(sum - arr[i]);
if (sum - (double)arr[i] == (double)arr[i])
twice_count--;
}
return twice_count / 2l;
}
static boolean[] sieveOfEratosthenes(int n)
{
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
prime[1]=false;
return prime;
}
static long power(long x, long y, long p)
{
long res = 1l;
x = x % p;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % p;
y>>=1;
x = (x * x) % p;
}
return res;
}
public static int log2(int N)
{
int result = (int)(Math.log(N) / Math.log(2));
return result;
}
////////////////////////////////////////////////////////////////////////////////////
////////////////////DO NOT READ AFTER THIS LINE //////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////
static long modFact(int n,
int p)
{
if (n >= p)
return 0;
long result = 1l;
for (int i = 2; i <= n; i++)
result = (result * i) % p;
return result;
}
static boolean isPalindrom(char[] arr, int i, int j) {
boolean ok = true;
while (i <= j) {
if (arr[i] != arr[j]) {
ok = false;
break;
}
i++;
j--;
}
return ok;
}
static int max(int a, int b) {
return Math.max(a, b);
}
static int min(int a, int b) {
return Math.min(a, b);
}
static long max(long a, long b) {
return Math.max(a, b);
}
static long min(long a, long b) {
return Math.min(a, b);
}
static int abs(int a) {
return Math.abs(a);
}
static long abs(long a) {
return Math.abs(a);
}
static void swap(long arr[], int i, int j) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int arr[], int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static int maxArr(int arr[]) {
int maxi = Integer.MIN_VALUE;
for (int x : arr)
maxi = max(maxi, x);
return maxi;
}
static int minArr(int arr[]) {
int mini = Integer.MAX_VALUE;
for (int x : arr)
mini = min(mini, x);
return mini;
}
static long maxArr(long arr[]) {
long maxi = Long.MIN_VALUE;
for (long x : arr)
maxi = max(maxi, x);
return maxi;
}
static long minArr(long arr[]) {
long mini = Long.MAX_VALUE;
for (long x : arr)
mini = min(mini, x);
return mini;
}
static int lcm(int a,int b){
return (int)(((long)a*b)/(long)gcd(a,b));
}
static long lcm(long a,long b){
return ((a*b)/(long)gcd(a,b));
}
static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static void ruffleSort(int[] a) {
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n);
int temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
Arrays.sort(a);
}
public static int binarySearch(int a[], int target) {
int left = 0;
int right = a.length - 1;
int mid = (left + right) / 2;
int i = 0;
while (left <= right) {
if (a[mid] <= target) {
i = mid + 1;
left = mid + 1;
} else {
right = mid - 1;
}
mid = (left + right) / 2;
}
return i-1;
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
int[][] read2dArray(int n, int m) {
int arr[][] = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
arr[i][j] = nextInt();
}
}
return arr;
}
ArrayList<Integer> readArrayList(int n) {
ArrayList<Integer> arr = new ArrayList<Integer>();
for (int i = 0; i < n; i++) {
int a = nextInt();
arr.add(a);
}
return arr;
}
long nextLong() {
return Long.parseLong(next());
}
}
static class SegmentTree {
int seg[];
int arr[];
int n;
public SegmentTree(int arr[], int n) {
this.n = n;
seg = new int[n * 4];
this.arr = arr;
}
int task(int a, int b) {
return Math.max(a, b);// TO DO
}
int mSeg(int i, int l, int r) {
if (l == r) {
seg[i] = task(arr[l], arr[l]);
return arr[l];
}
int m = (l + r) / 2;
return seg[i] = task(mSeg(2 * i + 1, l, m), mSeg(2 * i + 2, m + 1, r));
}
int qSeg(int ql, int qr, int i, int l, int r) {
if (r < ql || l > qr)
return Integer.MIN_VALUE;// TO DO
if (ql <= l && qr >= r)
return seg[i];
int m = (l + r) / 2;
return task(qSeg(ql, qr, 2 * i + 1, l, m), qSeg(ql, qr, 2 * i + 2, m + 1, r));
}
}
static class Pair{
int first, second,third;
Pair(int f, int s,int t){
first = f;
second = s;
third=t;
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | c6eab1bdec9e0f3fb68dc2797cb13d89 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class pchain {
public static void main(String[] args) throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(in.readLine());
for (int x = 0; x < t; x++) {
int n = Integer.parseInt(in.readLine());
System.out.println(n);
for (int i = 0; i < n; i++) {
int curr = 2;
StringBuilder sb = new StringBuilder();
for (int j = 0; j < n; j++) {
if (j == i) sb.append(1+" ");
else {
sb.append(curr+" ");
curr++;
}
}
System.out.println(sb.toString());
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | de1e7d363fc81dd94ec5ae1d84d5775e | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Stack;
public class Main {
static long MOD = 998244353l;
int min = Integer.MAX_VALUE;
int max = 0;
char result[][];
int count = 0;
int pattern = 0;
public static void main(String[] args) throws Exception {
// FileInputStream fis = new FileInputStream(new File("1.txt"));
var sc = new FastScanner();
// var sc = new FastScanner(fis);
// var pw = new FastPrintStream("test_normal_result.csv");
var pw = new FastPrintStream();
solve(sc, pw);
sc.close();
pw.flush();
pw.close();
}
public static void solve(FastScanner sc, FastPrintStream pw) {
int times = sc.nextInt();
for (int t = 0; t < times; t++) {
int n = sc.nextInt();
pw.println(n);
int a[] = new int[n];
for (int i=0;i<n;i++) {
a[i] = i+1;
}
for (int i=0;i<n;i++) {
for (int j=0;j<n;j++) {
pw.print(a[j]+" ");
}
pw.println();
if (i<n-1) {
int temp = a[n-i-1];
a[n-i-1] = a[n-i-2];
a[n-i-2] = temp;
}
}
}
}
public static Point checkPoint(char ch) {
Point re = new Point(0, 0);
switch (ch) {
case 'N': {
re.x--;
break;
}
case 'S': {
re.x++;
break;
}
case 'E': {
re.y++;
break;
}
case 'W': {
re.y--;
break;
}
}
return re;
}
public char[][] copyArray(char c[][], int n) {
char re[][] = new char[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
re[i][j] = c[i][j];
}
}
return re;
}
public static void swap(int[] s, int i, int j) {
int tmp = s[i];
s[i] = s[j];
s[j] = tmp;
}
public void permutation(int[] s, int from, int to, int a[]) {
if (to <= 1)
return;
if (from == to) {
check(s, a);
} else {
for (int i = from; i <= to; i++) {
swap(s, i, from);
permutation(s, from + 1, to, a);
swap(s, from, i);
}
}
}
public void check(int[] s, int a[]) {
int re = 0;
int b[][] = new int[2][3];
for (int i = 0; i < 6; i++) {
b[i / 3][i % 3] = a[s[i]];
}
int maxx = 0;
int maxy = 0;
int minx = 10000;
int miny = 10000;
for (int i = 0; i < 3; i++) {
maxx = Math.max(maxx, b[0][i]);
minx = Math.min(minx, b[0][i]);
maxy = Math.max(maxy, b[1][i]);
miny = Math.min(miny, b[1][i]);
}
re = (maxx - minx) * (maxy - miny) * 2;
re = re - (Math.abs(b[0][0] - b[0][1]) * Math.abs(b[1][0] - b[1][1]));
re = re - (Math.abs(b[0][0] - b[0][2]) * Math.abs(b[1][0] - b[1][2]));
re = re - (Math.abs(b[0][2] - b[0][1]) * Math.abs(b[1][2] - b[1][1]));
max = Math.max(re, max);
}
public static long anothertoTen(long ano, int another) {
long ten = 0;
long now = 1;
long temp = ano;
while (temp > 0) {
long i = temp % 10;
ten += now * i;
now *= another;
temp /= 10;
}
return ten;
}
public static long tentoAnother(long ten, int another) {
Stack<Long> stack = new Stack<Long>();
while (ten > 0) {
stack.add(ten % another);
ten /= another;
}
long re = 0;
while (!stack.isEmpty()) {
long pop = stack.pop();
re = re * 10 + pop;
}
return re;
}
// 2C5 = 5*4/(2*1)
public static long XCY(long x, long y) {
long temp = 1;
for (int i = 0; i < x; i++) {
temp = (temp * (y - i)) % MOD;
}
long tempx = 1;
for (int i = 2; i <= x; i++) {
tempx = (tempx * i) % MOD;
}
tempx = modpow(tempx, (long) MOD - 2);
temp = (temp * tempx) % MOD;
return temp;
}
static long modpow(long N, Long K) {
return BigInteger.valueOf(N).modPow(BigInteger.valueOf(K), BigInteger.valueOf(MOD)).longValue();
}
static long modpow(long N, Long K, long mod) {
return BigInteger.valueOf(N).modPow(BigInteger.valueOf(K), BigInteger.valueOf(mod)).longValue();
}
public static long gcd(long a, long b) {
if (b == 0) {
return a;
}
if (a < b) {
return gcd(b, a);
}
return gcd(b, a % b);
}
public static int gcd(int a, int b) {
if (b == 0) {
return a;
}
if (a < b) {
return gcd(b, a);
}
return gcd(b, a % b);
}
}
class Node implements Comparable<Node> {
int tyoten;
long minDistance;
public Node(int t, long m) {
tyoten = t;
minDistance = m;
}
@Override
public int compareTo(Node o) {
int res = -1;
if (this.minDistance - o.minDistance >= 0) {
res = 1;
}
return res;
}
}
class Range {
long x = 0;
long c = 0;
public Range(long x, long c) {
this.x = x;
this.c = c;
}
}
class Vertex {
String key;
Vertex(String key) {
this.key = key;
}
}
class Edge {
Vertex start;
Vertex end;
long key;
Edge(Vertex start, Vertex end, long key) {
this.start = start;
this.end = end;
this.key = key;
}
}
class Target {
int t;
long c;
boolean bool;
}
class Point implements Comparable {
int x;
int y;
public Point() {
}
public Point(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Object p) {
Point t = (Point) p;
if (this.x > t.x) {
return 1;
}
if (this.x < t.x) {
return -1;
}
return 0;
}
@Override
public boolean equals(Object p) {
Point t = (Point) p;
return this.x == t.x;
}
}
class PointY implements Comparable {
int a;
int b;
public int compareTo(Object p) {
PointY t = (PointY) p;
if (this.a > t.a) {
return -1;
}
if (this.a < t.a) {
return 1;
}
return 0;
}
}
class FastPrintStream implements AutoCloseable {
private static final int BUF_SIZE = 1 << 15;
private final byte[] buf = new byte[BUF_SIZE];
private int ptr = 0;
private final java.lang.reflect.Field strField;
private final java.nio.charset.CharsetEncoder encoder;
private java.io.OutputStream out;
public FastPrintStream(java.io.OutputStream out) {
this.out = out;
java.lang.reflect.Field f;
try {
f = java.lang.String.class.getDeclaredField("value");
f.setAccessible(true);
} catch (NoSuchFieldException | SecurityException e) {
f = null;
}
this.strField = f;
this.encoder = java.nio.charset.StandardCharsets.US_ASCII.newEncoder();
}
public FastPrintStream(java.io.File file) throws java.io.IOException {
this(new java.io.FileOutputStream(file));
}
public FastPrintStream(java.lang.String filename) throws java.io.IOException {
this(new java.io.File(filename));
}
public FastPrintStream() {
this(System.out);
try {
java.lang.reflect.Field f = java.io.PrintStream.class.getDeclaredField("autoFlush");
f.setAccessible(true);
f.set(System.out, false);
} catch (IllegalAccessException | IllegalArgumentException | NoSuchFieldException e) {
// ignore
}
}
public FastPrintStream println() {
if (ptr == BUF_SIZE)
internalFlush();
buf[ptr++] = (byte) '\n';
return this;
}
public FastPrintStream println(java.lang.Object o) {
return print(o).println();
}
public FastPrintStream println(java.lang.String s) {
return print(s).println();
}
public FastPrintStream println(char[] s) {
return print(s).println();
}
public FastPrintStream println(char c) {
return print(c).println();
}
public FastPrintStream println(int x) {
return print(x).println();
}
public FastPrintStream println(long x) {
return print(x).println();
}
public FastPrintStream println(double d, int precision) {
return print(d, precision).println();
}
private FastPrintStream print(byte[] bytes) {
int n = bytes.length;
if (ptr + n > BUF_SIZE) {
internalFlush();
try {
out.write(bytes);
} catch (java.io.IOException e) {
throw new RuntimeException();
}
} else {
System.arraycopy(bytes, 0, buf, ptr, n);
ptr += n;
}
return this;
}
public FastPrintStream print(java.lang.Object o) {
return print(o.toString());
}
public FastPrintStream print(java.lang.String s) {
if (strField == null) {
return print(s.getBytes());
} else {
try {
return print((byte[]) strField.get(s));
} catch (IllegalAccessException e) {
return print(s.getBytes());
}
}
}
public FastPrintStream print(char[] s) {
try {
return print(encoder.encode(java.nio.CharBuffer.wrap(s)).array());
} catch (java.nio.charset.CharacterCodingException e) {
byte[] bytes = new byte[s.length];
for (int i = 0; i < s.length; i++) {
bytes[i] = (byte) s[i];
}
return print(bytes);
}
}
public FastPrintStream print(char c) {
if (ptr == BUF_SIZE)
internalFlush();
buf[ptr++] = (byte) c;
return this;
}
public FastPrintStream print(int x) {
if (x == 0) {
if (ptr == BUF_SIZE)
internalFlush();
buf[ptr++] = '0';
return this;
}
int d = len(x);
if (ptr + d > BUF_SIZE)
internalFlush();
if (x < 0) {
buf[ptr++] = '-';
x = -x;
d--;
}
int j = ptr += d;
while (x > 0) {
buf[--j] = (byte) ('0' + (x % 10));
x /= 10;
}
return this;
}
public FastPrintStream print(long x) {
if (x == 0) {
if (ptr == BUF_SIZE)
internalFlush();
buf[ptr++] = '0';
return this;
}
int d = len(x);
if (ptr + d > BUF_SIZE)
internalFlush();
if (x < 0) {
buf[ptr++] = '-';
x = -x;
d--;
}
int j = ptr += d;
while (x > 0) {
buf[--j] = (byte) ('0' + (x % 10));
x /= 10;
}
return this;
}
public FastPrintStream print(double d, int precision) {
if (d < 0) {
print('-');
d = -d;
}
d += Math.pow(10, -d) / 2;
print((long) d).print('.');
d -= (long) d;
for (int i = 0; i < precision; i++) {
d *= 10;
print((int) d);
d -= (int) d;
}
return this;
}
private void internalFlush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (java.io.IOException e) {
throw new RuntimeException(e);
}
}
public void flush() {
try {
out.write(buf, 0, ptr);
out.flush();
ptr = 0;
} catch (java.io.IOException e) {
throw new RuntimeException(e);
}
}
public void close() {
try {
out.close();
} catch (java.io.IOException e) {
throw new RuntimeException(e);
}
}
private static int len(int x) {
int d = 1;
if (x >= 0) {
d = 0;
x = -x;
}
int p = -10;
for (int i = 1; i < 10; i++, p *= 10)
if (x > p)
return i + d;
return 10 + d;
}
private static int len(long x) {
int d = 1;
if (x >= 0) {
d = 0;
x = -x;
}
long p = -10;
for (int i = 1; i < 19; i++, p *= 10)
if (x > p)
return i + d;
return 19 + d;
}
}
class FastScanner implements AutoCloseable {
private final java.io.InputStream in;
private final byte[] buf = new byte[2048];
private int ptr = 0;
private int buflen = 0;
public FastScanner(java.io.InputStream in) {
this.in = in;
}
public FastScanner() {
this(System.in);
}
private boolean hasNextByte() {
if (ptr < buflen)
return true;
ptr = 0;
try {
buflen = in.read(buf);
} catch (java.io.IOException e) {
throw new RuntimeException(e);
}
return buflen > 0;
}
private int readByte() {
return hasNextByte() ? buf[ptr++] : -1;
}
public boolean hasNext() {
while (hasNextByte() && !(32 < buf[ptr] && buf[ptr] < 127))
ptr++;
return hasNextByte();
}
private StringBuilder nextSequence() {
if (!hasNext())
throw new java.util.NoSuchElementException();
StringBuilder sb = new StringBuilder();
for (int b = readByte(); 32 < b && b < 127; b = readByte()) {
sb.appendCodePoint(b);
}
return sb;
}
public String next() {
return nextSequence().toString();
}
public String next(int len) {
return new String(nextChars(len));
}
public char nextChar() {
if (!hasNextByte())
throw new java.util.NoSuchElementException();
return (char) readByte();
}
public char[] nextChars() {
StringBuilder sb = nextSequence();
int l = sb.length();
char[] dst = new char[l];
sb.getChars(0, l, dst, 0);
return dst;
}
public char[] nextChars(int len) {
if (!hasNext())
throw new java.util.NoSuchElementException();
char[] s = new char[len];
int i = 0;
int b = readByte();
while (32 < b && b < 127 && i < len) {
s[i++] = (char) b;
b = readByte();
}
if (i != len) {
throw new java.util.NoSuchElementException(
String.format("Next token has smaller length than expected.", len));
}
return s;
}
public long nextLong() {
if (!hasNext())
throw new java.util.NoSuchElementException();
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b)
throw new NumberFormatException();
while (true) {
if ('0' <= b && b <= '9') {
n = n * 10 + b - '0';
} else if (b == -1 || !(32 < b && b < 127)) {
return minus ? -n : n;
} else
throw new NumberFormatException();
b = readByte();
}
}
public int nextInt() {
return Math.toIntExact(nextLong());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public void close() {
try {
in.close();
} catch (java.io.IOException e) {
throw new RuntimeException(e);
}
}
}
/**
* @verified https://atcoder.jp/contests/practice2/tasks/practice2_j
*/
class SegTree<S> {
final int MAX;
final int N;
final java.util.function.BinaryOperator<S> op;
final S E;
final S[] data;
@SuppressWarnings("unchecked")
public SegTree(int n, java.util.function.BinaryOperator<S> op, S e) {
this.MAX = n;
int k = 1;
while (k < n)
k <<= 1;
this.N = k;
this.E = e;
this.op = op;
this.data = (S[]) new Object[N << 1];
java.util.Arrays.fill(data, E);
}
public SegTree(S[] dat, java.util.function.BinaryOperator<S> op, S e) {
this(dat.length, op, e);
build(dat);
}
private void build(S[] dat) {
int l = dat.length;
System.arraycopy(dat, 0, data, N, l);
for (int i = N - 1; i > 0; i--) {
data[i] = op.apply(data[i << 1 | 0], data[i << 1 | 1]);
}
}
public void set(int p, S x) {
exclusiveRangeCheck(p);
data[p += N] = x;
p >>= 1;
while (p > 0) {
data[p] = op.apply(data[p << 1 | 0], data[p << 1 | 1]);
p >>= 1;
}
}
public S get(int p) {
exclusiveRangeCheck(p);
return data[p + N];
}
public S prod(int l, int r) {
if (l > r) {
throw new IllegalArgumentException(String.format("Invalid range: [%d, %d)", l, r));
}
inclusiveRangeCheck(l);
inclusiveRangeCheck(r);
S sumLeft = E;
S sumRight = E;
l += N;
r += N;
while (l < r) {
if ((l & 1) == 1)
sumLeft = op.apply(sumLeft, data[l++]);
if ((r & 1) == 1)
sumRight = op.apply(data[--r], sumRight);
l >>= 1;
r >>= 1;
}
return op.apply(sumLeft, sumRight);
}
public S allProd() {
return data[1];
}
public int maxRight(int l, java.util.function.Predicate<S> f) {
inclusiveRangeCheck(l);
if (!f.test(E)) {
throw new IllegalArgumentException("Identity element must satisfy the condition.");
}
if (l == MAX)
return MAX;
l += N;
S sum = E;
do {
l >>= Long.numberOfTrailingZeros(l);
if (!f.test(op.apply(sum, data[l]))) {
while (l < N) {
l = l << 1;
if (f.test(op.apply(sum, data[l]))) {
sum = op.apply(sum, data[l]);
l++;
}
}
return l - N;
}
sum = op.apply(sum, data[l]);
l++;
} while ((l & -l) != l);
return MAX;
}
public int minLeft(int r, java.util.function.Predicate<S> f) {
inclusiveRangeCheck(r);
if (!f.test(E)) {
throw new IllegalArgumentException("Identity element must satisfy the condition.");
}
if (r == 0)
return 0;
r += N;
S sum = E;
do {
r--;
while (r > 1 && (r & 1) == 1)
r >>= 1;
if (!f.test(op.apply(data[r], sum))) {
while (r < N) {
r = r << 1 | 1;
if (f.test(op.apply(data[r], sum))) {
sum = op.apply(data[r], sum);
r--;
}
}
return r + 1 - N;
}
sum = op.apply(data[r], sum);
} while ((r & -r) != r);
return 0;
}
private void exclusiveRangeCheck(int p) {
if (p < 0 || p >= MAX) {
throw new IndexOutOfBoundsException(
String.format("Index %d out of bounds for the range [%d, %d).", p, 0, MAX));
}
}
private void inclusiveRangeCheck(int p) {
if (p < 0 || p > MAX) {
throw new IndexOutOfBoundsException(
String.format("Index %d out of bounds for the range [%d, %d].", p, 0, MAX));
}
}
// **************** DEBUG **************** //
private int indent = 6;
public void setIndent(int newIndent) {
this.indent = newIndent;
}
@Override
public String toString() {
return toSimpleString();
}
public String toDetailedString() {
return toDetailedString(1, 0);
}
private String toDetailedString(int k, int sp) {
if (k >= N)
return indent(sp) + data[k];
String s = "";
s += toDetailedString(k << 1 | 1, sp + indent);
s += "\n";
s += indent(sp) + data[k];
s += "\n";
s += toDetailedString(k << 1 | 0, sp + indent);
return s;
}
private static String indent(int n) {
StringBuilder sb = new StringBuilder();
while (n-- > 0)
sb.append(' ');
return sb.toString();
}
public String toSimpleString() {
StringBuilder sb = new StringBuilder();
sb.append('[');
for (int i = 0; i < N; i++) {
sb.append(data[i + N]);
if (i < N - 1)
sb.append(',').append(' ');
}
sb.append(']');
return sb.toString();
}
}
class DSU {
private int n;
private int[] parentOrSize;
public DSU(int n) {
this.n = n;
this.parentOrSize = new int[n];
Arrays.fill(parentOrSize, -1);
}
int merge(int a, int b) {
if (!(0 <= a && a < n) || !(0 <= b && b < n)) {
return -1;
}
int x = leader(a);
int y = leader(b);
if (x == y)
return x;
if (-parentOrSize[x] < -parentOrSize[y]) {
int tmp = x;
x = y;
y = tmp;
}
parentOrSize[x] += parentOrSize[y];
parentOrSize[y] = x;
return x;
}
boolean same(int a, int b) {
if (!(0 <= a && a < n) || !(0 <= b && b < n)) {
return false;
}
return leader(a) == leader(b);
}
int leader(int a) {
if (parentOrSize[a] < 0) {
return a;
} else {
parentOrSize[a] = leader(parentOrSize[a]);
return parentOrSize[a];
}
}
int size(int a) {
if (!(0 <= a && a < n)) {
return -1;
}
return -parentOrSize[leader(a)];
}
ArrayList<ArrayList<Integer>> groups() {
int[] leaderBuf = new int[n];
int[] groupSize = new int[n];
for (int i = 0; i < n; i++) {
leaderBuf[i] = leader(i);
groupSize[leaderBuf[i]]++;
}
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
for (int i = 0; i < n; i++) {
result.add(new ArrayList<>());
}
for (int i = 0; i < n; i++) {
result.get(leaderBuf[i]).add(i);
}
return result;
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 308ef656495b0eaf85f83cf2890455fa | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Test
{
final static private FastReader fr = new FastReader();
final static private PrintWriter out = new PrintWriter(System.out) ;
final static private long mod = (long)1e9 + 7;
private static void solve()
{
int n = fr.nextInt() ;
int[] arr = new int[n] ;
for (int i = 0 ; i < n; i++)
{
arr[i] = i+1 ;
}
out.println(n);
for (int x: arr)
out.print(x+" ");
out.println();
for (int j = n-1 ; j > 0; j--)
{
int temp = arr[j] ;
arr[j] = arr[j-1] ;
arr[j-1] = temp ;
for (int x: arr)
out.print(x+" ");
out.println();
}
}
public static void main(String[] args)
{
int t = 1 ;
t = fr.nextInt() ;
while (t-- > 0)
{
solve() ;
}
out.close() ;
}
private static long[] inputArr(int n){
long[] arr = new long[n] ;
for (int i = 0 ;i < n ; i++) {
arr[i] = fr.nextLong() ;
}
return arr;
}
private static void sort(int[] arr){
ArrayList<Integer> al = new ArrayList<>() ;
for (int x : arr){
al.add(x) ;
}
Collections.sort(al);
for (int i = 0 ; i < arr.length; i++){
arr[i] = al.get(i) ;
}
}
private static void sort(long[] arr){
ArrayList<Long> al = new ArrayList<>() ;
for (long x : arr){
al.add(x) ;
}
Collections.sort(al);
for (int i = 0 ; i < arr.length; i++){
arr[i] = al.get(i) ;
}
}
private static long getMax(long ... a) {
long max = Long.MIN_VALUE ;
for (long x : a) max = Math.max(x, max) ;
return max ;
}
private static long getMin(long ... a) {
long max = Long.MAX_VALUE ;
for (long x : a) max = Math.min(x, max) ;
return max ;
}
private static long fastPower(double a, long b) {
double ans = 1 ;
while (b > 0)
{
if ((b & 1) != 0) ans *= a ;
a *= a ;
b >>= 1 ;
}
return (long)(ans + 0.5) ;
}
private static long fastPower(long a, long b, long mod) {
long ans = 1 ;
while (b > 0)
{
if ((b&1) != 0) ans = (ans%mod * a%mod) %mod;
b >>= 1 ;
a = (a%mod * a%mod)%mod ;
}
return ans ;
}
private static int lower_bound(List<Integer> arr, int key) {
int pos = Collections.binarySearch(arr, key) ;
if (pos < 0) {
pos = - (pos + 1) ;
}
return pos ;
}
private static int upper_bound(List<Integer> arr, int key) {
int pos = Collections.binarySearch(arr, key);
pos++ ;
if (pos < 0) {
pos = -(pos) ;
}
return pos ;
}
private static int upper_bound(int[] arr, int key) {
int start = 0 , end = arr.length ;
while (start < end) {
int mid = start + ((end - start) >> 1) ;
if (arr[mid] <= key) start = mid + 1 ;
else end = mid ;
}
return start ;
}
private static int lower_bound(int[] arr, int key) {
int start = 0 , end = arr.length;
while (start < end) {
int mid = start + ((end - start )>>1) ;
if (arr[mid] >= key){
end = mid ;
}
else start = mid + 1 ;
}
return start ;
}
private static class Pair{
int x ;
int y ;
Pair(int x, int y){
this.x = x ;
this.y = y ;
}
@Override
public int hashCode() {
return super.hashCode();
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair pair = (Pair) o;
return x == pair.x && y == pair.y;
}
}
private static long gcd(long a, long b) {
if (b == 0) return a ;
return gcd(b, a%b) ;
}
private static long lcm(long a, long b) {
return (a * b)/gcd(a, b);
}
private static boolean isPrime(int num) {
if (num <= 2) return true ;
for (int i = 2; i <= Math.sqrt(num); i++)
{
if (num%i == 0) return false ;
}
return true;
}
private static List<Long> seive(int n) {
// all are false by default
// false -> prime, true -> composite
boolean[] nums = new boolean[n+1] ;
for (int i = 2 ; i <= Math.sqrt(n); i++) {
if (!nums[i]) {
for (int j = i*i ; j <= n ; j += i) {
nums[j] = true ;
}
}
}
ArrayList<Long>primes = new ArrayList<>() ;
for (int i = 2 ; i <=n ; i++) {
if (!nums[i]) primes.add((long) i) ;
}
return primes ;
}
private static boolean isVowel(char ch)
{
return ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u';
}
private static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | e11b6ae8164595347cb1e31d73571f79 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.*;
import java.util.*;
public class gotoJapan {
public static void main(String[] args) throws java.lang.Exception {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solution solver = new Solution();
boolean isTest = true;
int tC = isTest ? Integer.parseInt(in.next()) : 1;
for (int i = 1; i <= tC; i++)
solver.solve(in, out, i);
out.close();
}
/* ............................................................. */
static class Solution {
InputReader in;
PrintWriter out;
public void solve(InputReader in, PrintWriter out, int test) {
this.in = in;
this.out = out;
int n=ni();
int x=1;
pn(n);
for(int i=1;i<=n;i++) {
int p=2;
for(int j=1;j<=n;j++) {
if(x==j)out.print(1+" ");
else {
out.print(p+" ");
p++;
}
}
x++;
out.println();
}
}
class Pair {
int x;
int y;
long w;
Pair(int x, int y, long w) {
this.x = x;
this.y = y;
this.w = w;
}
}
char[] n() {
return in.next().toCharArray();
}
int ni() {
return in.nextInt();
}
long nl() {
return in.nextLong();
}
long[] nal(int n) {
long a[] = new long[n];
for (int i = 0; i < n; i++)
a[i] = nl();
return a;
}
void pn(long zx) {
out.println(zx);
}
void pn(String sz) {
out.println(sz);
}
void pn(double dx) {
out.println(dx);
}
void pn(long ar[]) {
for (int i = 0; i < ar.length; i++)
out.print(ar[i] + " ");
out.println();
}
void pn(String ar[]) {
for (int i = 0; i < ar.length; i++)
out.println(ar[i]);
}
}
/* ......................Just Input............................. */
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
/* ......................Just Input............................. */
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 92fd588f7db4c670d1d38ee7444d07bd | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class A {
static class Pair {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
static long gcd(long n, long m) {
if (m == 0)
return n;
else
return gcd(m, n % m);
}
static long lcm(long n, long m) {
return (n * m) / gcd(n, m);
}
static void dfs(ArrayList<ArrayList<Integer>> adj, int s, int[] vis, int[] cnt) {
for (int i : adj.get(s)) {
if (vis[i] == 1) {
continue;
}
cnt[0]++;
vis[i] = 1;
dfs(adj, i, vis, cnt);
}
}
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
StringBuilder str = new StringBuilder();
int t = sc.nextInt();
for (int xx = 0; xx < t; xx++) {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = i + 1;
str.append(n+"\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
str.append(arr[j] + " ");
}
if (i != n - 1) {
int temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
}
str.append("\n");
}
}
System.out.println(str);
sc.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 11 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 7d4d6bf2a8104bfc92e8fec096df8c2e | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.util.stream.IntStream;
import static java.lang.System.out;
import static java.util.Arrays.stream;
public class B {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t-->0) {
int n = scanner.nextInt();
int[] a = new int[n];
for (int i = 1; i<=n; i++) {
a[i-1]+=i;
}
out.println(n);
printArray(a);
for (int i = n-1; i>=1; i--) {
swap(a, i, i-1);
printArray(a);
}
}
}
private static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
private static void printArray(int[] a) {
StringBuilder builder = new StringBuilder();
for (int i : a)
builder.append(i).append(" ");
builder.append("\n");
out.print(builder);
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | dc734cdb3a3ebc5aea0fcee60f06e5b5 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.util.stream.IntStream;
import static java.lang.System.out;
import static java.util.Arrays.stream;
public class B {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t-->0) {
int n = scanner.nextInt();
out.println(n);
for (int i = 1; i<=n; i++) {
StringBuilder builder = new StringBuilder(i+" ");
for (int j = 1; j<=n; j++) {
if (j==i) continue;
builder.append(j).append(" ");
}
out.println(builder);
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 6480e1cfec24464c994f61d7e7e5e313 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.BufferedWriter;
import java.io.OutputStreamWriter;
import java.util.Scanner;
import java.io.IOException;
import java.util.Arrays;
import java.util.ArrayList;
public class Main{
public static void main(String[] args) throws IOException {
Scanner sc= new Scanner(System.in);
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
int t=sc.nextInt();
while(t>0){
t-=1;
int n=sc.nextInt();
int a[]=new int[n];
for(int i=0; i<n; i++) a[i]=i+1;
int y=n-1;
writer.write(n + "\n");
for(int i=0; i<n; i++){
for(int j=0; j<n; j++){
writer.write(a[j] + " ");
}
writer.write("\n");
int temp=a[i];
a[i]=a[y];
a[y]=temp;
}
}
sc.close();
writer.flush();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 72d8d08c4493d6bc6127437429462041 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution extends Helper {
private static final SuperFastReader sc = new SuperFastReader();
// private static final FastReader sc = new FastReader();
// private static final Scanner sc = new Scanner(System.in);
private static final PrintWriter out = new PrintWriter(System.out);
private static final int MOD = (int) (1e9) + 7;// ((a + b) % MOD + MOD) % MOD
public static void main(String[] args) throws IOException {
int t = sc.Int();
for (int i = 1; i <= t; ++i) {
// System.out.print("Case #" + i + ": ");
solve();
}
// solve();
sc.close();
out.close();
}
public static void solve() throws IOException {
int n = sc.Int();
int perm[] = new int[n];
println(n);
for (int i = 0; i < n; i++) {
perm[i] = i + 1;
print(perm[i] + " ");
}
println();
for (int i = 2; i < n; i++) {
int t = perm[i];
perm[i] = perm[i - 1];
perm[i - 1] = t;
for (int j = 0; j < n; j++) {
print(perm[j] + " ");
}
println();
}
int t = perm[n - 1];
perm[n - 1] = perm[0];
perm[0] = t;
for (int j = 0; j < n; j++) {
print(perm[j] + " ");
}
println();
}
public static <T> void print(T data) {
out.print(data);
}
public static <T> void println(T data) {
out.println(data);
}
public static void println() {
out.println();
}
public static void read(int arr[]) throws IOException {
for (int i = 0; i < arr.length; i++) {
arr[i] = sc.Int();
}
}
public static void read(long arr[]) throws IOException {
for (int i = 0; i < arr.length; i++) {
arr[i] = sc.Long();
}
}
public static void read(String arr[]) throws IOException {
for (int i = 0; i < arr.length; i++) {
arr[i] = sc.next();
}
}
public static void read(int mat[][]) throws IOException {
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat[i].length; j++) {
mat[i][j] = sc.Int();
}
}
}
public static void read(long mat[][]) throws IOException {
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat[i].length; j++) {
mat[i][j] = sc.Long();
}
}
}
}
class Pair<K, V> {
public K key;
public V value;
Pair(K k, V v) {
key = k;
value = v;
}
}
class SuperFastReader {
private InputStream stream;
private byte[] buf = new byte[1 << 16];
private int curChar, numChars;
// standard input
public SuperFastReader() {
stream = System.in;
}
// file input
public SuperFastReader(String i) throws IOException {
stream = new FileInputStream(i);
}
private int nextByte() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars == -1)
return -1; // end of file
}
return buf[curChar++];
}
public String next() {
int c;
do {
c = nextByte();
} while (c <= ' ');
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = nextByte();
} while (c > ' ');
return res.toString();
}
public String nextLine() {
int c;
do {
c = nextByte();
} while (c < '\n');
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = nextByte();
} while (c > '\n');
return res.toString().trim(); // .trim() used to remove '\n' from either ends
}
public int Int() {
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public long Long() {
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public double Double() {
return Double.parseDouble(next());
}
public char Char() {
return next().charAt(0);
}
public void close() throws IOException{
stream.close();
}
}
class Helper {
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 564726ae8b8ec7d3cd1438de4c5f069b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution {
private static final SuperFastReader sc = new SuperFastReader();
private static final int MOD = (int) (1e9) + 7;// ((a + b) % MOD + MOD) % MOD
public static void main(String[] args) throws IOException {
int t = sc.Int();
for (int i = 1; i <= t; ++i) {
// System.out.print("Case #" + i + ": ");
solve();
}
// solve();
sc.close();
}
public static void solve() throws IOException {
int n = sc.Int();
int perm[] = new int[n];
StringBuilder res = new StringBuilder();
res.append(n).append("\n");
for (int i = 0; i < n; i++) {
perm[i] = i + 1;
res.append(perm[i] + " ");
}
res.append("\n");
for (int i = 2; i < n; i++) {
int t = perm[i];
perm[i] = perm[i - 1];
perm[i - 1] = t;
for (int j = 0; j < n; j++) {
res.append(perm[j] + " ");
}
res.append("\n");
}
int t = perm[n - 1];
perm[n - 1] = perm[0];
perm[0] = t;
for (int j = 0; j < n; j++) {
res.append(perm[j] + " ");
}
res.append("\n");
System.out.print(res);
}
}
class Pair<K, V> {
public K key;
public V value;
Pair(K k, V v) {
key = k;
value = v;
}
}
class SuperFastReader extends PrintWriter {
private InputStream stream;
private byte[] buf = new byte[1 << 16];
private int curChar, numChars;
// standard input
public SuperFastReader() {
this(System.in, System.out);
}
public SuperFastReader(InputStream i, OutputStream o) {
super(o);
stream = i;
}
// file input
public SuperFastReader(String i, String o) throws IOException {
super(new FileWriter(o));
stream = new FileInputStream(i);
}
private int nextByte() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars == -1)
return -1; // end of file
}
return buf[curChar++];
}
public String next() {
int c;
do {
c = nextByte();
} while (c <= ' ');
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = nextByte();
} while (c > ' ');
return res.toString();
}
public String nextLine() {
int c;
do {
c = nextByte();
} while (c < '\n');
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = nextByte();
} while (c > '\n');
return res.toString().trim(); // .trim() used to remove '\n' from either ends
}
public int Int() {
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public long Long() {
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public double Double() {
return Double.parseDouble(next());
}
public char Char(){
return next().charAt(0);
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | d2284a52960741f6f3a61bba39dfbccb | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class CF1716B_build_permutation {
// For fast input output
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
try {
br = new BufferedReader(
new FileReader("input.txt"));
PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
System.setOut(out);
} catch (Exception e) {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
// end of fast i/o code
public static boolean isPalindrome(String str) {
int i = 0;
int j = str.length() - 1;
int flag = 0;
while (i <= j) {
if (str.charAt(i) != str.charAt(j)) {
flag = 1;
break;
}
i++;
j--;
}
return flag == 1 ? false : true;
}
public static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
class Pair {
int x1;
int x2;
public Pair(int x1, int x2) {
this.x1 = x1;
this.x2 = x2;
}
}
static void buildPermut(int n ,ArrayList<Integer>temp,int i,boolean vis[],ArrayList<ArrayList<Integer>>ans){
if(temp.size()==n){
ArrayList<Integer>t=new ArrayList<>();
for(int j:temp)
t.add(j);
ans.add(t);
// System.out.println(ans);
return;
}
for(int j =1;j<=n;j++){
if(!vis[j]){
vis[j]=true;
temp.add(j);
buildPermut(n, temp, j, vis, ans);
temp.remove(temp.size()-1);
vis[j]=false;
}
}
}
static void swap(int nums[],int i,int j){
int temp=nums[i];
nums[i]=nums[j];
nums[j]=temp;
}
public static void main(String[] args) {
FastReader fs = new FastReader();
PrintWriter out= new PrintWriter(System.out);
int t=fs.nextInt();
while(t-->0){
int n =fs.nextInt();
out.println(n);
int ans []=new int[n+1];
for(int i =1;i<=n;i++){
out.print(i+" ");
ans[i]=i;
}
out.println();
for(int i =1;i<n;i++){
swap(ans,i,i+1);
for(int j =1;j<=n;j++)
out.print(ans[j]+" ");
out.println();
}
}
out.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | f65203134e3bb7d38dc1fc6c5beba2ba | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class InVoker {
//Variables
static long mod = 1000000007;
static long mod2 = 998244353;
static FastReader inp= new FastReader();
static PrintWriter out= new PrintWriter(System.out);
public static void main(String args[]) {
InVoker g=new InVoker();
g.main();
out.close();
}
//Main
void main() {
// Ara ara :/
int t=inp.nextInt();
loop:
while(t-->0) {
int n=inp.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++) {
a[i]=i+1;
}
out.println(n);
print(a);
a[0]=a[n-1];
a[n-1]=1;
print(a);
for(int j=0;j<n-2;j++) {
int x=a[j];
a[j]=a[j+1];
a[j+1]=x;
print(a);
}
}
}
/*********************************************************************************************************************************************************************************************************
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*ti;. .:::;: fDEEEEEEKKKKKKKKKEKEEEEEEEEEEEEEEE#WEEEEEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKW#KKKKKKKKKKEEf .:::. .,;tfLGDEDDDDDDDDEEE *
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*tti. ....::::.. ...,. fEEEKKKKKKKKKKKKKKKW#WWWWWWWWWWWWWWWWtiEiiiWWWKKEWKEi....D.EDDDEi;.fWEEEEEEEEEEDDfL.;EDDEEEWEEEEEEEEEEEEWWEEEDDDDDGf. :;ijfLGGDDDDDDDDDDDD *
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*jti. ...:. ,EEKKKKKWWKKKKKWWWWWWWWW###WWWWWWWWKKKKfWWLt;i,. fi EG..D:.............EKEKK;;..t....:.......KWKWWWWKKKKKKKWKK####WKKKWEEEDf:. .,;tfGGGGDDDDDDDDDDD *
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*jti. ...:. GEKKKKKWWKKKKKWWWWWWWWWW##WWWWWWWKKKKKKt;EiKKKK, ...t................jEKKG;;..,.....,LGi....KWWWWWWKKKKKKWKKKW####WKKKKKEEL,,,:. .,;tfGGGDDDDDDDDDDDD *
*jti. ...:. .GEEKKKKKWKKKKKWWWWWWWWWW###WWWWWWWKKKKKKtiE::tGG........................EEEj;;...,.........:D..DKWWWWWWKKKKK#KKW###W#WKKKKKEEfj:,,,:. .,;tfGGGDDDDDDDDDDDD *
*jti. ...:. DEKKKKKWWKKKKKWWWWWWWWW####WWWWWWWKKKKKKiiE:::.::.......................EEi;;...j.....f......:iDKWWWWWWKKKKK#WW######WKKKKKEELG :,,,,:. .,;tfGGGDDDDDDDDDDDD *
*jti. ...:. fEEKKKKWWKKKKWWWWWWWWWWW###WWWWWWWWKKKKKK;tE::..........................DD;.;,.::......;........EWWWWWWWKKKKW#WW#####WWKKKWKKELG .:,,,:::,;tfGGGDDDDDDDDDDDD *
*jti. ...:. .DEKEKKKWWKKKKWWWWWWWWWWW###WWWWWWWWKKKKKE,iD::..........................D..,;.,;tLffi...........DWDWWWW#KKKWWWWW#####W#KKKWKEEGL .:,;,,,;tfGGGDDDDDDDDDDDD *
*jti. ...:. ;EEKKKKWWKKKKKWWWWWW#WWWW####WWWWWWKKKKKEL:iD:..........................j ..;..;;:.....i,........DKtWWWWWKKWWWWWW#####WWWKKKEKEDf .:,;;;itfGGGDDDDDDDDDDDD *
*jti. ...:. DEKKKKKWWKKKKWWWWWWW#WWWW####WWWWWWKKKKKEj:iG...............................:....................GKiWWWWWKKWW#WWW######WWKKKKKEEf .,;iitfGGGDDDDDDDDDDDD *
*jti. ...:.:EKKKKKWWKKKKKWWWWWWW#WWW#####WWWWWKWKKKKEi:if:.................................iEKEKKKKKKDj......DKiWWWWWKWK##WW#######WWKKK:KEEL .:;itfGGGDDDDDDDDDDDD *
*jji. ...:,DEEKKKWWWKWKKWWWWWWWWWWWW#####WWWWWWWKKKKEi:it..................................j. KKKKKKKKKKKf..DKiWWWWWKWW##WW#######WWKKK,KEEf .,;tfGGGDDDDDDDDDDDD *
*jji. ..L:iDEEKKKWWKKKKKWWWWWWWWWWWW#####WWWWWKWKKKKKi.i;.................................. . KKKWWWWWWWWK..DGiWWWWWKK##WWW#####W#WWKKKjEKEL, .:;tfGGGDDDDDDDDDDDD *
*jji. .f:::EEEKKKWWWKKKKKWWWWWWWWWWWW#####WWWWWKWKKKKK;.i,.................................:: KKEKWWWWWWfWK..EiiWWWWWKWW#WW##########KKKD,KELj .:;tfGGDDDDDDDDDDDDD *
*jji. .t::::,DEEKKKWWKKKKWWWWWWWWW#WWWW#####WWWWKKWKKKEK;.i:.................................GDDEEEKKKWWWWWtWWD.E;iWWWWWW###WW#########WWKKK.EEDG .:;tfGGGDDDDDDDDDDDD *
*jji. . j..::::EKEKKKWWWKKKKWWWWWWWWW#WWW######WWWWKKWKKKEK;.t:.................................ELLEDDEEEWWWWEtWK,.KiiWWWWWW###W##########WWKKK:EEEG .;tjfLLGDDDDDDDDDDDDDDD *
*jji. i.::::::,EEEKKWWWKKKKKWWWWWWWWW#WWW#####WWWWWKWKKKKEE,.t..................................DfiEGDDDEEKKKttKWG.KiiWWWWW##WWW##########WWKKK:fEEL ,fGGGDDDDDDDDEEEDDDDDDDDDD *
*jji. .;:..:::::DEEEKKWWWKKKKKWWWWWWWWW#WWWW####WWWWWWWKKKKED,.t..................................ifjDDGGEGDKK.ttKKE.DiWWWWW###WW##########WWWKKK:.KELiLGGGGDDDDDDDDDDDDEEEDDDDDDD *
*jji. i.:.::::::,KEEKKWWWKKKKKKWWWWWWWWW#WWWW####WWWWWWWKKKKEL:.j..................................GGf,;ifLLED .iiKKi:fWWWWWW##W#W##########WWWKKK:.KKLGGDDDDDDDDDDDDDDDDEDDEEDDDDD *
*jji. .j:.::::::::EEEKKKWWWKKKKKKWWWWWWWW##WWW#####WWWWWWWKKKKKf:.f..................................:EEfftf .,. ;iE,..jWWWWWWW###W############WWKKK,:KKGDDDDDDDDDDDDDDDDDDDDDDDEDDDD *
*jji. .:.::::::::,,EEEKKWWWKKKKKKKWWWWWWWW##WWW#####WWWWWWWKKKKKt..G....................................EEELL; .j....tKWWWWWWW################WWWKKtfGKGEDDDDDDDDDDDDDDDDDDDDDDDEEDD *
*jji. :...:::::::,,jEEKKWWWWKKKKKKWWWWWWWWW##KWW#####KWWWWWWKKKKEi..D....................................:jEEE.........;KKWWWWWWWW#WW##W##########WWKKDLGKEKDDDDDDDDDDDDDDDDDDDDDDDDDED *
*jji. i:.::::::::,,,EEEKKWWWWKKKKKWWWWWWWWWW##WWW#####WWWWWWWKKKKKi..D......................................:::::......,KKKWWWWWWWWW#####W########WWWKKKGGKKEGGGGGGGGDDDDDDDDDDDDDDDDDDE *
*jji. i..:::::::::,,tEEKKWWWWKKKKKWWWWWWWWWWW##WW######WWWWWWWKKKKKi..D......................................::::......:EKKKWWWWWWWWWWW##WW########W#WKKWGGKKGGGGGGGGGGGGGGGDDDDDDDDDDDDD *
*jji. .:::::::::::,,,EEEKKWWWWKKKKKWWWWWWWWWWW##WW#####WWWWWWWWKKKKKi..D....................................:::::::::..tELii;KWWWWWWWWWW##WW######WWWWWWKWGGGKGGGGGGGGGGGGGGGGGGGGGGGGGGDG *
*jjt. :.::::::::,,,,fEEKKWWWWKKKKKKWWWWWWWWWW###WW####WWWWWWW#WKKKKKi..D....................................:::::::.:.,;;;;;;,KKWWWWWWWWW#WW########WWWKKWGGGKGGGGGGGGGGGGGGGGGGGGGGGGGGGG *
*jji. ;.::::::::,,,,;EEEKWWWWWKKKKKWWWWWWWWWWWW##WW###WKWWWWWK#WKKKKKi..G......................................:::::::,;;;;:...KKKWWWWWWWWWKWW#######WWWWKKGLGKDGGGGGGLLGGGGGGGGGGGGGGGGGGG *
*jjt. f.:::::::::,,,,fEEKKWWWWWKKKKKWWWWWWWWWWW###WW##WKKWWWWWW#WKKKKK;.jt........i.............................:::::::;j;;....:E.KKKWWWWWWWKWW#####W#WWWWKKLLGWEEGGGGGLGGGGGGGGGGGGGGGGGGGG *
*jjt. ...:::::::,,,,,;DEEKWWWWWKKKKKWWWWWWWWWWWW####WWWKKKWWWWWWWWKKKKK;.E;.........t.............................:::::ii;;.....D...KKWWWWWWWKWW#####WWEWWWKKGGGEKKGGGGGLGGGGGGGGGGGGGLGGGGGG *
*fji. ;.:::::::,,,,,;LEEKKWWWWWKKKKKWWWWWWWWWWWW####KWKKKKWWWWWWWWKKKKKi.D;..........j.............................:::tt;,.....:.....KKWWWWWWKWWWW##WWWGWWWKKGGGGKEGGGGGGGGGGGGGGGGGGGLLGGGGL *
*fji. t::::::::,,,,,,;EEEKWWWWWKKKKKKKWWWWWWWWWWW##WKWKKKKKWWWWWWWWKKKKKi:D;............j...........................::LL;,.............KKWWWWWKWWWWWWWWWGWWWKKGGGGKGGGGGGGGGGGGGGGGGGGGLLGGGGL *
*fjt: .:::::::,,,,,,,DEEKWWWWWWKKKKKKKWWWWWWWWWWWWKKWKKKKKKWWWWK#WWKKKKWitE;........... ............................:G;;:...............KKKWWKKWWWWWWWWWGWWWKKGGGGWGGGGGGGGGGGGGGGGGGGGGGGGGGL *
*fjji;:. .f:::::::,,,,,,,;EEEKWWWWWWKKKKKKWWWWWWWWWWWKKKKKKKKKKKWWKWWWWWKKKKWGKD;........................................L;;..................DKKWKKWWWWWWWWWGWWWKKDGGGKDGGGGGGGGGGGGGGGGGGGGGGGGGG *
*fjjtii;,:. :::::::,,,,,,,;EEEKWWWWWWKKKKKKWWWWWWWWWWKKKKKKKKKKKKWWWWWW#WWKKKKWiEj;......................................:i,;....,...............;KKEKWWWWWWWWWGKWWKKDDGGDEGGGDGGGGGDGGGGGGGGGGGGGGGG *
*fjtiiiii;;:. j::::::,,,,,,,;;EEEKWWWWW#KKKKKWWWWWWWWWKKKKKKWKKKKKKKWWWWWWWWWKKKKWtEL;:....................................;;;:...,;j................:KEEWWWWWWWWWDDWWKKDDDDDKDDDDDDDDDDDDDDDGGGGGGGGGGG *
*fjti;;iiii;;,:::::::,,,,,,,,;EEEKWWWWWWWKKKKWWWWWWWWKKKKKKKWKKKKKKKWWWWWWW#W#KKKKWEEii;...................................f;:....,;L...................EEKWWWWWWWWDDWWKKDDDDDKEDDDDDDDDDDDDDDDDDDDDDGGGG *
*fjt,,,;;;;ii;f::::::,,,,,,,;;EEKWWWWWWWKKKKKWWWKWWKKKKKKKKKKKKKKKKKWWWWWWW#W#KKKKWKEij;:...............................:G;,.....,;f....................:tKKWWWWWWWDDWWKKDDDDDKKDDDDDDDDDDDDDDDDDDDDDDDDD *
*jjt. ..:,;;;;,::::,,,,,,,,;;GEEWWWWWWWWKKKKWKKWKKKKKKKKKKKKKKKKKKKKWWWWWWW#W#KKKKWEDi;j;............................,Li;L;;;..,;;f........................KKKKWWWKDDWWKKDDDGDKKGGGGGGGGDGDDDDDDDDDDDDDDD *
*fjt. .:,,,:::::,,,,,,,;;;EEKWWWWWWWKKKKKKWKKKKKKKKKKKKKKKKKKKKKWKKKWKW##W#KKKKWEti;;G;........................tEEEL;;;;;;;;;;L..........................DKKKKKEDDWWKEDGftiLE;;;;itjLGGGGGGDDDDDDDDDDD *
*fjt. .j::::,,,,,,,;;;DEEWWWWWWWWKKKKWKKKKKKKKKKKKKKKKKKKKKKKWKKWWWK##W#KKKKKEii;;;L;...................iDEEEEEEKKi;j;;;;jD.....:......................,KKKKDGGEKKE:::::;E::::::::::,tLGGDDDDDDDDDD *
*fjt. .;:::,,,,,,,;;;;EEKWWWWWWWWKWKKKKKKKKKKKKKKKWKKKKKKKKKKWKKWWWW#WW#KKKKKKii;;;;f;.............:tDEEEEEKKKKKKKKEti;;;L...............................EEKf;:iKKE::::::E::::::::::::::ifDDDDDDDDD *
*fjt: :::,,,,,,,,;;;DEEWWWWWWWWWEKKKKKKKKKKKKKKKKKKKKKKKKKKWWKKWWWW####KKKKKEiii;;;;f,.........iDEEEEKKKKKKKKKKKKKKKf;iG......i..........................fK::::KKE::::::E::::::::::::::::,tGGDDDDD *
*fjt: t:::,,,,,,;;;;iDEKWWWWWWKEKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWKKWWWW####WKKKKLiii;;;;;L,....,Li;EDEEEEKKKKKKKKKKKKKKKKiG......;:...........................:i:::KKE:::::,E,::::::::::::::::::iGDDDD *
*jjt. f::,,,,,,,;;;;GEEWWWWKEEKEKKKKKKKKKKKKKKKKWKKKKKKKKKKKWWKWWWWW###WWKKKKiii;;;;;;;G,;L;;iiEEEEEEEKKKKKKKKKKKKKWWKE......;t.........:....................j::KEE:,::,,D,,::::,,,,,,:::::::::tDDD *
*fjt:. ,::,,,,,,,;;;;EEWWKEEEEEEKKKKKKKKKKKKKKKKWKKKKKKKKKKKWWKKWWWWW#W#KKKKKKiiiiii;;;;;i;;iiiEEKEEKKWKKKKKKKWKKKKKWWWGi;...;t......,;;;;,....................:,EEE,,,,,,D,,,,,,,,,,,,,,,,::,::::tG *
*fjt:. ,::,,,,,,,;;;;DEKEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWWW#W#KKKKKKiiiii;;i;;;;;iiiKKKEKKKKWWKWWWWWWKKKKWWWWW;;;:;L.....;;;;;;;;;....................,KEE,,,,,,E,,,,,,,,,,,,,,,,,,,,,,,,; *
*fjt:. f:,,,,,,,;;;;jEDEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWW#W##KKKKKKiiiiiiii;i;;iiiEKKKKKKKKWKWWWWWWWWKKKWWWWWKi;;i.....,jEEfGi;;;;;...................EED,,,,,,E,,,,,,,,,,,,,,,,,,,,,,,,, *
*fjt:. .f::,,,,,,;;jEEDEEEEEEEEEEKKKKKKKKKKKKKKKWKKKKKKKKKKKKKWWWKWWWWW###KKKKKLiiiiiiiiiiiiiiEEKKKKKKKKWWWWWWWWWWWWKWWWWWWGi;i;,..;jDDDKEGi;;;;;;:................EED,,,,,,D,,,,,,,,,,,,,,,,,,,,,,,,, *
*fjt:. .. ;::,,,,,;;EDDEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWKKW#WW####KWKKKiiiiiiiiiiiiijKKKKKKKKKKWWWWWWWWWWWWWWWWWWWWWt;i;;;;i;DDDDDDGi;;;;;;;;:.............EDf;,,,;,G;;;;;;;;;;;;;;;,,,,,,,,,, *
*fjt:......:,,,,,,;LDDDEEEEEEEEEEEKKKKKKKKKKKKKKKKWKKKKKKKKKKKKKWWWWKWWWW####KKKKKiiiiiiiiiiijKEKKWKKKKKKKWWWWWWWWWWWWWWWWWWWWWWiLiii;i;DEEEEDDE;i;;;;;;;;;:..........EDi,;;;;;L;;;;;;;;;;;;;;;;;;,,,,,,, *
*fjt:......:,,,,,;EDDDEEKEEEEEEEEEKKKKKKKKKKKKKKKWKKKKKKKKKKKKKKWWWWKKWWW##W#KWKKWEiiiiiijGKKKKKWWKKKKKKKKWWWWWWWWWWWWWWWWWWWWWWKi;iiiiDDEEEEEEDEi;;;;;;;;;;;;;,:.....ED;;;;;;;j;;;;;;;;;;;;;;;;;;;;;;;,, *
*fjt:.....t,,,,,;DDDDEEEKEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWKKKWWWW##WKWKKWKiiiKKKKKKKKKWWKKKKKKKKWWWWWWWWWWWWWWW#WWWWWWWWWiiiiifLEEEEEEEEDi;i;;;;;;;;;;;;.....DD;;;;;;;i;;;;;;;;;;;;;;;;;;;;;;;;; *
*fjt:.....G,,,,,GDDDEEEEEEEEEEEEKKKKKKKKKKKKKKKKWKKKKKKKKKKKKKKKWWWKKKWWW###WKWKKWKitKKKKKKKKKWKKKKKKKKKKWWWWWWWWWWWWWW###WWWWWWWWEiiiiiiiEEEEEEEEDGiiii;;;;;;;;;.....GD;;;;;;;i;;;;;;;;;;;;;;;;;;;;;;;;; *
*fjt:.....L,,,,;GDDDEEEEEEEEEEKEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWDGWWW###KKWWKWKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWWWWWWW####WWWWWWWWWiiiiiiiiEEEEEEEEEEDi;i;;;;;;;;.....Lj;;;;;;i;iiiiii;;;;;;ii;;;;;;;;;;; *
***********************************************************************************************************************************************************************************************************/
class Edge {
int to;
long a,b;
Edge(int to, long a, long b){
this.to=to;
this.a=a;
this.b=b;
}
}
void sort(int a[]) {
ArrayList<Integer> list=new ArrayList<>();
for(int x: a) list.add(x);
Collections.sort(list);
for(int i=0;i<a.length;i++) a[i]=list.get(i);
}
void sort(long a[]) {
ArrayList<Long> list=new ArrayList<>();
for(long x: a) list.add(x);
Collections.sort(list);
for(int i=0;i<a.length;i++) a[i]=list.get(i);
}
void ruffleSort(int a[]) {
Random rand=new Random();
int n=a.length;
for(int i=0;i<n;i++) {
int j=rand.nextInt(n);
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
Arrays.sort(a);
}
void ruffleSort(long a[]) {
Random rand=new Random();
int n=a.length;
for(int i=0;i<n;i++) {
int j=rand.nextInt(n);
long temp=a[i];
a[i]=a[j];
a[j]=temp;
}
Arrays.sort(a);
}
static long gcd(long a, long b) {
return b==0?a:gcd(b,a%b);
}
long fact[];
long invFact[];
void init(int n) {
fact=new long[n+1];
invFact=new long[n+1];
fact[0]=1;
for(int i=1;i<=n;i++) {
fact[i]=mul(i,fact[i-1]);
}
invFact[n]=power(fact[n],mod-2);
for(int i=n-1;i>=0;i--) {
invFact[i]=mul(invFact[i+1],i+1);
}
}
long nCr(int n, int r) {
if(n<r || r<0) return 0;
return mul(fact[n],mul(invFact[r],invFact[n-r]));
}
long add(long a, long b) {
return (a+b)%mod;
}
long mul(long a, long b) {
return a*b%mod;
}
long power(long x, long y) {
long gg=1;
while(y>0) {
if(y%2==1) gg=mul(gg,x);
x=mul(x,x);
y/=2;
}
return gg;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st==null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String s="";
try {
s=br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return s;
}
}
// Functions
static int gcd(int a, int b) {
return b==0?a:gcd(b,a%b);
}
void print(int a[]) {
int n=a.length;
for(int i=0;i<n;i++) out.print(a[i]+" ");
out.println();
}
void print(long a[]) {
int n=a.length;
for(int i=0;i<n;i++) out.print(a[i]+" ");
}
//Input Arrays
static void input(long a[], int n) {
for(int i=0;i<n;i++) {
a[i]=inp.nextLong();
}
}
static void input(int a[], int n) {
for(int i=0;i<n;i++) {
a[i]=inp.nextInt();
}
}
static void input(String s[],int n) {
for(int i=0;i<n;i++) {
s[i]=inp.next();
}
}
static void input(int a[][], int n, int m) {
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
a[i][j]=inp.nextInt();
}
}
}
static void input(long a[][], int n, int m) {
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
a[i][j]=inp.nextLong();
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 13034a22d7c63715be28c7a5cee9c47b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
public class ACMP {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
System.out.println(n);
for (int i = 1; i <= n; i++) {
StringBuilder builder = new StringBuilder(i + " ");
for (int j = 1; j <= n; j++) {
if (j == i) continue;
builder.append(j).append(" ");
}
System.out.println(builder);
}
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 8ea0c06bce9b8ea64d2ce8b3f9a12b08 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.PrintWriter;
public class PermutationChain {
public static void main(String args[]) {
Scanner sc =new Scanner(System.in);
int tc = sc.nextInt();
PrintWriter out = new PrintWriter(System.out);
int[] arr = new int[101];
while(tc != 0){
tc--;
int n = sc.nextInt();
for(int i = 0; i < n ; i++){
arr[i] = i+1;
}
int l = n-2;
out.println(n);
for(int i = 0; i < n; i++){
out.print(arr[i] + " ");
}
out.println();
while(l >= 0){
int temp = arr[l];
arr[l] = arr[l+1];
arr[l+1] = temp;
l--;
for(int i = 0; i < n; i++){
out.print(arr[i] + " ");
}
out.println();
}
out.flush();
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | a2cc062df9bdb83840a1f8ec2e973baa | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import javax.print.DocFlavor.STRING;
import java.io.*;
import java.lang.Math;
public class cp{
public static PrintWriter out;
public static FastReader sc;
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void sort(int[] a,boolean flag){
ArrayList<Integer> temp=new ArrayList<>();
for(int n:a) temp.add(n);
Collections.sort(temp);
if(flag){
for(int i=0;i<a.length;i++){
a[i]=temp.get(i);
}
}
else{
for(int i=a.length-1;i>=0;i--){
a[i]=temp.get(i);
}
}
}
public static int nod(int n){
return (int)Math.log10((double)n)+1;
}
public static int gcd(int a,int b){
if (b == 0)
return a;
else
return gcd(b, a % b);
}
public static void printarr(int[] a){
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printlist(List<Integer> a){
for(int i:a){
out.print(i+" ");
}
out.println();
}
public static boolean isPrime(long n)
{
if(n < 2) return false;
if(n == 2 || n == 3) return true;
if(n%2 == 0 || n%3 == 0) return false;
long sqrtN = (long)Math.sqrt(n)+1;
for(long i = 6L; i <= sqrtN; i += 6) {
if(n%(i-1) == 0 || n%(i+1) == 0) return false;
}
return true;
}
public static HashMap<Integer, Integer> sortMap(HashMap<Integer, Integer> hm)
{
List<Map.Entry<Integer, Integer> > list = new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
public static HashMap<Integer,Integer> mapFreq(int[] a){
HashMap<Integer,Integer> map=new HashMap<>();
for(int i=0;i<a.length;i++){
map.put(a[i],map.getOrDefault(a[i], 0)+1);
}
return map;
}
public static void solve(){
// int n=sc.nextInt();
// int[] a=new int[n];
// for(int i=0;i<n;i++){
// a[i]=sc.nextInt();
// }
int n=sc.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++) a[i]=i+1;
out.println(n);
for(int i=n-1;i>=1;i--){
printarr(a);
int temp=a[i];
a[i]=a[i-1];
a[i-1]=temp;
}
printarr(a);
}
public static void main(String[] args) throws IOException {
sc = new FastReader();
out = new PrintWriter(new BufferedOutputStream(System.out));
int t=sc.nextInt();
// int t=1;
while (t-->0){
solve();
}
out.close();
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 102a9219cde4bd76364a1377c69efdbc | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.io.PrintWriter;
import java.util.*;
public class PermutationChain {
public static void main(String[] args) {
int t,n;
Scanner sc=new Scanner(System.in);
PrintWriter out=new PrintWriter(System.out);
t=sc.nextInt();
for (int g=0; g<t; g++){
n=sc.nextInt();
out.println(n);
int[] arr;
arr=new int[n];
for (int h=0; h<n; h++){
arr[h]=h+1;
out.print(arr[h]+" ");
}
out.println();
for (int q=1; q<n-1; q++){
int w=arr[q-1];
arr[q-1]=arr[n-1];
arr[n-1]=w;
for (int s=0; s<n; s++){
out.print(arr[s]+" ");
}
out.println();
}
int e=arr[n-2];
arr[n-2]=arr[n-1];
arr[n-1]=e;
for (int p=0; p<n; p++){
out.print(arr[p]+" ");
}
out.println();
}
out.flush();
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | df3aa5bc0936a9623855537d0006b13b | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.PrintWriter;
public class MyClass {
public static void main(String args[]) {
Scanner sc =new Scanner(System.in);
int tc = sc.nextInt();
PrintWriter out = new PrintWriter(System.out);
int[] arr = new int[101];
while(tc != 0){
tc--;
int n = sc.nextInt();
for(int i = 0; i < n ; i++){
arr[i] = i+1;
}
int l = n-2;
out.println(n);
for(int i = 0; i < n; i++){
out.print(arr[i] + " ");
}
out.println();
while(l >= 0){
int temp = arr[l];
arr[l] = arr[l+1];
arr[l+1] = temp;
l--;
for(int i = 0; i < n; i++){
out.print(arr[i] + " ");
}
out.println();
}
out.flush();
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 1e28e35d8d814b5c318a6eff3fb1fff1 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes |
import java.util.*;
public class Test1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
System.out.println(n);
int[] arr = new int[n+1];
StringBuilder sb = new StringBuilder();
for(int i=1;i<=n;i++) {
arr[i] = i;
sb.append(i).append(" ");
// System.out.print(i +" ");
}
System.out.println(sb.toString());
sb.setLength(0);
arr[1] = n;
arr[n] = 1;
int k = n-1;
int j = 1;
while(k-->0) {
for(int i=1;i<=n;i++)
sb.append(arr[i]).append(" ");
// System.out.print(arr[i] +" ");
System.out.println(sb.toString());
sb.setLength(0);
arr[j] = arr[j+1];
arr[j+1] = n;
j++;
}
}
}
}
| Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 6b3b1148da82b4aeab53fe2fc4f0147e | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main{
public static int[] array(BufferedReader br,int n) throws IOException{
String [] values = br.readLine().split(" ");
int [] arr = new int[n];
for(int i =0; i<n; i++){
arr[i] = Integer.parseInt(values[i]);
}
return arr;
}
public static void swap(int[] arr, int x, int y){
int temp = arr[x];
arr[x] = arr[y];
arr[y] = temp;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0){
int n = Integer.parseInt(br.readLine());
int [] arr = new int[n];
StringBuilder sb = new StringBuilder();
for(int i = 0; i<n; i++){
arr[i] = i+1;
}
// System.out.println(n);
sb.append(n +"\n");
for(int i = 0; i<n; i++){
//System.out.print(arr[i]+ " ");
sb.append(arr[i] + " ");
}
//System.out.println();
sb.append("\n");
for(int i = 0; i<n-1; i++){
swap(arr,i,n-1);
for(int j = 0; j<n; j++){
// System.out.print(arr[j]+ " ");
sb.append(arr[j] + " ");
}
// System.out.println();
sb.append("\n");
}
System.out.print(sb);
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 5f68d784421d70cdc5d46f9d85df3d76 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | /**/
import java.io.*;
import java.util.*;
import java.lang.*;
public class PermutationChain {
public static void main(String[] args) throws IOException{
FastReader s = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = s.nextInt();
while(t-->0){
int n = s.nextInt();
int cur = 0;
out.println(n);
for(int i = 0; i < n; i++){
StringBuilder sb = new StringBuilder();
for(int j = 0; j < n - 1; j++){
if(j == cur){
sb.append(1);
sb.append(' ');
}
sb.append(j + 2);
sb.append(' ');
}
cur++;
if(i == n - 1){
sb.append(1);
}
out.println(sb);
}
}
out.flush();
}
public static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public String next() throws IOException {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
long nextLong() throws IOException {
return Long.parseLong(next());
}
double nextDouble() throws IOException {
return Double.parseDouble(next());
}
String nextLine() throws IOException {
String str = "";
if (st.hasMoreTokens()) {
str = st.nextToken("\n");
} else {
str = br.readLine();
}
return str;
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output | |
PASSED | 45e90878a7e805a351f71712928286c2 | train_109.jsonl | 1659623700 | A permutation of length $$$n$$$ is a sequence of integers from $$$1$$$ to $$$n$$$ such that each integer appears in it exactly once.Let the fixedness of a permutation $$$p$$$ be the number of fixed points in it — the number of positions $$$j$$$ such that $$$p_j = j$$$, where $$$p_j$$$ is the $$$j$$$-th element of the permutation $$$p$$$.You are asked to build a sequence of permutations $$$a_1, a_2, \dots$$$, starting from the identity permutation (permutation $$$a_1 = [1, 2, \dots, n]$$$). Let's call it a permutation chain. Thus, $$$a_i$$$ is the $$$i$$$-th permutation of length $$$n$$$.For every $$$i$$$ from $$$2$$$ onwards, the permutation $$$a_i$$$ should be obtained from the permutation $$$a_{i-1}$$$ by swapping any two elements in it (not necessarily neighboring). The fixedness of the permutation $$$a_i$$$ should be strictly lower than the fixedness of the permutation $$$a_{i-1}$$$.Consider some chains for $$$n = 3$$$: $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$ — that is a valid chain of length $$$2$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. $$$a_1 = [2, 1, 3]$$$, $$$a_2 = [3, 1, 2]$$$ — that is not a valid chain. The first permutation should always be $$$[1, 2, 3]$$$ for $$$n = 3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [1, 3, 2]$$$, $$$a_3 = [1, 2, 3]$$$ — that is not a valid chain. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped but the fixedness increase from $$$1$$$ to $$$3$$$. $$$a_1 = [1, 2, 3]$$$, $$$a_2 = [3, 2, 1]$$$, $$$a_3 = [3, 1, 2]$$$ — that is a valid chain of length $$$3$$$. From $$$a_1$$$ to $$$a_2$$$, the elements on positions $$$1$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$3$$$ to $$$1$$$. From $$$a_2$$$ to $$$a_3$$$, the elements on positions $$$2$$$ and $$$3$$$ get swapped, the fixedness decrease from $$$1$$$ to $$$0$$$. Find the longest permutation chain. If there are multiple longest answers, print any of them. | 256 megabytes | import static java.lang.System.out;
import static java.lang.Math.abs;
import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.log;
import java.util.*;
import java.lang.*;
import java.io.*;
public class a_Codeforces {
public static void main(String[] args) throws java.lang.Exception {
FastReader sc = new FastReader();
FastWriter out = new FastWriter();
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
out.println(n);
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = i + 1;
out.print((i + 1) + " ");
}
out.println("");
int i = 0, j = n - 1;
while (i < j) {
swap(arr, i, j);
i++;
for (int x : arr) {
out.print(x + " ");
}
out.println("");
}
}
out.close();
}
static void no() {
out.println("NO");
}
static void yes() {
out.println("YES");
}
static void print(int arr[]) {
for (int i = 0; i < arr.length; i++) {
out.print(arr[i] + " ");
}
out.println("");
}
static int[] input(int n) {
FastReader sc = new FastReader();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
return arr;
}
static int[] swap(int arr[], int a, int b) {
int temp = arr[b];
arr[b] = arr[a];
arr[a] = temp;
return arr;
}
/*
* int[] arr = new int[n];
* for (int i=0; i<n; i++){
* arr[i] = sc.nextInt();
* }
*/
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
} | Java | ["2\n\n2\n\n3"] | 2 seconds | ["2\n1 2\n2 1\n3\n1 2 3\n3 2 1\n3 1 2"] | null | Java 17 | standard input | [
"constructive algorithms",
"math"
] | 02bdd12987af6e666d4283f075f73725 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 99$$$) — the number of testcases. The only line of each testcase contains a single integer $$$n$$$ ($$$2 \le n \le 100$$$) — the required length of permutations in the chain. | 800 | For each testcase, first, print the length of a permutation chain $$$k$$$. Then print $$$k$$$ permutations $$$a_1, a_2, \dots, a_k$$$. $$$a_1$$$ should be an identity permutation of length $$$n$$$ ($$$[1, 2, \dots, n]$$$). For each $$$i$$$ from $$$2$$$ to $$$k$$$, $$$a_i$$$ should be obtained by swapping two elements in $$$a_{i-1}$$$. It should also have a strictly lower fixedness than $$$a_{i-1}$$$. | standard output |
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