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stringlengths 2
88
| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
values | programming_language
stringclasses 5
values | solution
stringlengths 1
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1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastReader in = new FastReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
DMinimumEulerCycle solver = new DMinimumEulerCycle();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class DMinimumEulerCycle {
public void solve(int testNumber, FastReader s, PrintWriter w) {
int n = s.nextInt();
long l = s.nextLong(), r = s.nextLong();
ArrayList<Integer> sizes = new ArrayList<>();
int cur = n - 1 << 1;
for (int i = n - 1; i > 0; i--) {
sizes.add(cur);
cur -= 2;
}
sizes.add(1);
cur = 0;
int dif = (int) (r - l + 1);
while (l > sizes.get(cur))
l -= sizes.get(cur++);
int[] ans = new int[2 * n];
if (cur == sizes.size() - 1) {
w.println(1);
return;
} else {
int cnt = cur + 1;
for (int i = 0; i < sizes.get(cur); i += 2) ans[i] = cnt;
cnt++;
for (int i = 1; i < sizes.get(cur); i += 2) ans[i] = cnt++;
}
for (int i = (int) l - 1; i < sizes.get(cur) && dif > 0; i++) {
w.print(ans[i] + " ");
dif--;
}
cur++;
while (dif > 0) {
if (cur == sizes.size() - 1) {
w.println(1);
return;
} else {
int cnt = cur + 1;
for (int i = 0; i < sizes.get(cur); i += 2) ans[i] = cnt;
cnt++;
for (int i = 1; i < sizes.get(cur); i += 2) ans[i] = cnt++;
}
for (int i = 0; i < sizes.get(cur) && dif > 0; i++) {
w.print(ans[i] + " ");
dif--;
}
cur++;
}
w.println();
}
}
static class FastReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private FastReader.SpaceCharFilter filter;
public FastReader(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 nextInt() {
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 long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long 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 next() {
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 interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
//BigInteger A;
//A= BigInteger.valueOf(54);
//ArrayList<Integer> a=new ArrayList<>();
//TreeSet<Integer> ts=new TreeSet<>();
//HashMap<Integer,Integer> hm=new HashMap<>();
//PriorityQueue<Integer> pq=new PriorityQueue<>();
public final class Practice
{
static ArrayList<Integer> dfsorder;
static int child[];
//static int c=0;
public static void dfs(ArrayList<ArrayList<Integer>> adj,boolean vis[],int u)
{
vis[u]=true;
dfsorder.add(u);
child[u]=1;
for(int i=0;i<adj.get(u).size();i++)
{
if(!vis[adj.get(u).get(i)])
{
dfs(adj,vis,adj.get(u).get(i));
child[u]+=child[adj.get(u).get(i)];
}
}
}
public static void main(String[]args)throws IOException
{
int K=(int)Math.pow(10,9)+7;
FastReader ob=new FastReader();
int t=ob.nextInt();
while(t-->0)
{
long n=ob.nextLong();
long l=ob.nextLong();
long r=ob.nextLong();
ArrayList<Long> a=new ArrayList<>();
long k=0;
while(l>2*(n-1)&&n>0)
{
l-=2*(n-1);
r-=2*(n-1);
++k;
--n;
}
while(l<=2*(n-1)&&n>0)
{
for(long i=l;i<=Math.min(r,2*n-2);i++)
{
if(i%2!=0)
a.add(1+k);
else
a.add((i/2)+1+k);
}
l=1;
r-=2*(n-1);
++k;
--n;
}
if(r==1)
a.add(1l);
for(int i=0;i<a.size();i++)
System.out.print(a.get(i)+" ");
System.out.println();
}
}
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine()
{
String s="";
try {
s=br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return s;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pr = pair<int, int>;
template <typename T>
void _read(T *arr, int n) {
for (int i = 0; i < n; i++) cin >> arr[i];
}
template <typename T>
void _write(T *arr, int n) {
for (int i = 0; i < n; i++) cout << arr[i] << " ";
cout << endl;
}
const int MAXN = 100033;
ll n;
ll l, r;
int start;
void __Main__() {
cin >> n >> l >> r;
ll sum = 0;
ll now = 2 * n - 2;
if (l == r && l == n * (n - 1) + 1) {
cout << 1 << endl;
return;
}
ll st = 0;
for (int i = 1; i < n; i++) {
if (sum + now >= l) {
st = sum;
start = i;
break;
}
sum += now;
now -= 2;
}
ll res = now;
ll i;
for (i = sum + 1;; i++) {
res--;
ll output;
if ((i - st) & 1) {
output = start;
} else {
output = start + (i - st) / 2;
}
if (res == 0) {
now -= 2;
res = now;
start++;
st = i;
}
if (i < l) {
continue;
}
if (i > r) break;
cout << output << " ";
if (i >= n * (n - 1)) break;
}
if (i < r) cout << 1;
cout << endl;
}
signed main() {
ios ::sync_with_stdio(false);
cin.tie(nullptr);
{
int _Test_cases;
cin >> _Test_cases;
for (int case_num = 1; case_num <= _Test_cases; case_num++) {
__Main__();
}
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.awt.*;
import java.util.List;
public class Main {
static int mod = (int) 1e9 + 7;
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
StringBuilder sb = new StringBuilder();
int t = sc.nextInt();
while(t-->0) {
long n = sc.nextLong(), l = sc.nextLong(), r = sc.nextLong();
long base, nonBase, startIndex;
base = (long) Math.floor(l / (2 * n));
while (((base * 2 * n) - (base * (base + 1)) < l) && (base < n)) base++;
startIndex = ((base - 1) * 2 * n) - (base * (base - 1));
for (long j = l; j < r + 1; j++) {
if (j > ((base * 2 * n) - (base * (base + 1)))) {
startIndex = j;
base += 1;
}
nonBase = base + (j - startIndex + 1) / 2;
if (j == (n * (n - 1)) + 1) {
sb.append(1);
break;
}
if (j % 2 == 1)sb.append(base+" ");
else sb.append(nonBase+" ");
}
sb.append("\n");
}
System.out.println(sb);
}
}
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;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
from sys import stdin
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def lsi():
x=list(stdin.readline())
x.pop()
return x
def si(): return stdin.readline()
res=['YES', 'NO']
############# CODE STARTS HERE #############
for _ in range(ii()):
n, l, r=mi()
if l==n*(n-1)+1:
print(1)
continue
x, p=1, n*2
while x<=l:
p-=2
x+=p
#print(x, p)
x-=p
p=n*2-p
p//=2
#print(x, p)
a=[]
dl=l-x+1
z=p+((dl+1)//2)
if not dl%2:
a.append(z)
l+=1
z+=1
if z>n:
p+=1
z=p+1
#print(a)
#print(p, z)
while l<=r:
a.append(p)
l+=1
if l<=r:
a.append(z)
l+=1
z+=1
if z>n:
p+=1
z=p+1
if r==n*(n-1)+1:
a[-1]=1
print(*a)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
for _ in range(int(input())):
n,l,r = map(int,input().split())
s=0
ans=[]
for i in range(1,n):
gap2=i
t=i
if s+(n-i)*2 >=l :
for j in range(l,r+1):
gg=(j-s)//2
if j%2==1:
ans.append(gap2)
else:
ans.append(gap2+gg)
if gap2+gg==n:
gap2+=1
s+=(n-t)*2
t+=1
if r==n*(n-1)+1:
ans[-1]=1
break
else:
s+=(n-i)*2
if len(ans)==0:
print(1)
else:
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long n, l, r;
void solve(long long l, long long r) {
long long s = 1;
while (s <= n && l > 2 * (n - s)) {
l -= 2 * (n - s);
r -= 2 * (n - s);
s++;
}
long long cnt = l / 2 + l % 2;
cnt += s;
while (s <= n && l <= r) {
while (cnt <= n) {
if (l % 2)
cout << s << " ";
else
cout << cnt++ << " ";
l++;
if (l > r) break;
}
s++;
cnt = s + 1;
}
if (l <= r) cout << 1;
cout << "\n";
}
signed main() {
long long t;
cin >> t;
while (t--) {
cin >> n >> l >> r;
solve(l, r);
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]) {new Main().run();}
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
void run(){
for(int q=ni();q>0;q--) {
work();
out.println();
}
out.flush();
}
long mod=998244353L;
long inf=Long.MAX_VALUE;
long gcd(long a,long b) {
return a==0?b:gcd(b%a,a);
}
void work() {
long n=nl(),l=nl(),r=nl();
long c=1;
long sum=0;
while(sum+(n-c)*2<l&&c<n) {
sum+=(n-c)*2;
c++;
}
// System.out.println(c);
long c2=(l+1-sum)/2+c;
if((l-sum)%2==0) {
out.print(c2+" ");
l++;
c2++;
if(c2>n) {
c++;
c2=c+1;
}
}
while(l<=Math.min(n*(n-1), r)) {
out.print(c+" ");
l++;
if(l>r)break;
out.print(c2+" ");
c2++;
if(c2>n) {
c++;
c2=c+1;
}
l++;
}
if(l<=r) {
out.print(1+" ");
}
}
//input
@SuppressWarnings("unused")
private ArrayList<Integer>[] ng(int n, int m) {
ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
for(int i=0;i<n;i++) {
graph[i]=new ArrayList<>();
}
for(int i=1;i<=m;i++) {
int s=in.nextInt()-1,e=in.nextInt()-1;
graph[s].add(e);
graph[e].add(s);
}
return graph;
}
private ArrayList<long[]>[] ngw(int n, int m) {
ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
for(int i=0;i<n;i++) {
graph[i]=new ArrayList<>();
}
for(int i=1;i<=m;i++) {
long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
graph[(int)s].add(new long[] {e,w,i});
graph[(int)e].add(new long[] {s,w});
}
return graph;
}
private int ni() {
return in.nextInt();
}
private long nl() {
return in.nextLong();
}
private String ns() {
return in.next();
}
private long[] na(int n) {
long[] A=new long[n];
for(int i=0;i<n;i++) {
A[i]=in.nextLong();
}
return A;
}
private int[] nia(int n) {
int[] A=new int[n];
for(int i=0;i<n;i++) {
A[i]=in.nextInt();
}
return A;
}
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br=new BufferedReader(new InputStreamReader(System.in));
}
public String next()
{
while(st==null || !st.hasMoreElements())//ε车οΌη©Ίθ‘ζ
ε΅
{
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt()
{
return Integer.parseInt(next());
}
public long nextLong()
{
return Long.parseLong(next());
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.concurrent.TimeUnit;
public class d {
public static void main(String[] args) throws IOException {
//FastReader scan = new FastReader("in.txt");
FastReader scan = new FastReader();
//PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("out.txt")));
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
Task solver = new Task();
int t = scan.nextInt();
//int t = 1;
for(int i = 1; i <= t; i++) solver.solve(i, scan, out);
out.close();
}
static class Task {
public void solve(int testNumber, FastReader sc, PrintWriter out) {
long N = sc.nextLong();
long L = sc.nextLong();
long R = sc.nextLong();
TreeMap<Long, Integer> div = new TreeMap<>();
div.put(0l, 1);
long change = (N-1) * 2;
int j = 2;
for(long i = 0; change > 0; j++) {
i += change;
div.put(i, j);
change -= 2;
}
//System.out.println(div);
long count = div.lowerKey(L);
while(count <= R) {
//System.out.println(count);
if(count == div.lastKey() + 1) {
print(out, count, L, R, 1);
} else if (div.containsKey(count)) {
print(out, count, L, R, N);
} else {
int repeat = div.get(div.lowerKey(count));
for(int i = repeat + 1; i < N; i++) {
print(out, count, L, R, repeat);
count++;
print(out, count, L, R, i);
count++;
}
print(out, count, L, R, repeat);
}
count++;
}
out.println();
}
public static void print(PrintWriter out, long count, long L, long R, long value) {
if(count >= L && count <= R)
out.print(value + " ");
}
}
static class tup implements Comparable<tup>, Comparator<tup> {
int a, b;
tup() {
}
tup(int a, int b) {
this.a = a;
this.b = b;
}
@Override
public int compareTo(tup o2) {
return a==o2.a?Integer.compare(b,o2.b):Integer.compare(a, o2.a);
}
@Override
public int compare(tup o1, tup o2) {
return o1.a==o2.a ? Integer.compare(o1.b, o2.b): Integer.compare(o1.a, o2.a);
}
@Override
public int hashCode() {
return Objects.hash(a, b);
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
tup other = (tup) obj;
return a==other.a && b==other.b;
}
}
static void shuffle(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
}
static void shuffle(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws FileNotFoundException {
br = new BufferedReader(new FileReader(new File(s)));
}
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;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastReader in = new FastReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
DMinimumEulerCycle solver = new DMinimumEulerCycle();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class DMinimumEulerCycle {
public void solve(int testNumber, FastReader s, PrintWriter w) {
int n = s.nextInt();
long l = s.nextLong(), r = s.nextLong();
ArrayList<Integer> sizes = new ArrayList<>();
int cur = n - 1 << 1;
for (int i = n - 1; i > 0; i--) {
sizes.add(cur);
cur -= 2;
}
sizes.add(1);
cur = 0;
while (l > sizes.get(cur)) {
l -= sizes.get(cur);
r -= sizes.get(cur++);
}
long dif = r - l + 1;
int[] ans = new int[4 * n];
if (cur == sizes.size() - 1) {
w.println(1);
return;
} else {
int cnt = cur + 1;
for (int i = 0; i < sizes.get(cur); i += 2) ans[i] = cnt;
cnt++;
for (int i = 1; i < sizes.get(cur); i += 2) ans[i] = cnt++;
}
for (int i = (int) l - 1; i < sizes.get(cur) && dif > 0; i++) {
w.print(ans[i] + " ");
dif--;
}
cur++;
while (dif > 0) {
if (cur == sizes.size() - 1) {
w.println(1);
return;
} else {
int cnt = cur + 1;
for (int i = 0; i < sizes.get(cur); i += 2) ans[i] = cnt;
cnt++;
for (int i = 1; i < sizes.get(cur); i += 2) ans[i] = cnt++;
}
for (int i = 0; i < sizes.get(cur) && dif > 0; i++) {
w.print(ans[i] + " ");
dif--;
}
cur++;
}
w.println();
}
}
static class FastReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private FastReader.SpaceCharFilter filter;
public FastReader(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 nextInt() {
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 long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long 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 next() {
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 interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie();
int T;
cin >> T;
while (T--) {
long long n, l, r;
cin >> n >> l >> r;
long long e = 0, s = -1, f = -1;
for (long long i = 1; i < n; i++) {
e += 2 * (n - i);
if (e >= l) {
s = i;
f = 2 * (n - i) - e + l;
break;
}
}
if (s == -1) {
cout << 1 << '\n';
continue;
}
for (long long i = l; i <= r; i++) {
if (s == n)
cout << 1 << ' ';
else {
if (f & 1)
cout << s << ' ';
else
cout << s + f / 2 << ' ';
}
if (f == 2 * (n - s)) {
s++;
f = 1;
} else
f++;
}
cout << '\n';
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
template <class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
using namespace std;
int main(void) {
cin.tie(0);
ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
long long l, r;
cin >> l >> r;
int idx = 0;
long long sum = 0;
int lg = -1, rg = -1;
int ls = -1, rs = -1;
while ((lg == -1 || rg == -1) && idx < n - 1) {
long long lsum = sum;
sum += (long long)(n - 1 - idx) * 2;
if (lg == -1 && l <= sum) lg = idx, ls = l - lsum;
if (rg == -1 && r <= sum) rg = idx, rs = r - lsum;
idx++;
}
if (lg == -1) lg = n - 1;
if (rg == -1) rg = n - 1;
lg++;
rg++;
for (int i = (int)(lg); i <= (int)(rg); i++) {
if (i == n) {
cout << "1 ";
break;
}
long long from = 1, to = (long long)(n - i) * 2;
if (i == lg) from = ls;
if (i == rg) to = rs;
for (int j = (int)(from); j <= (int)(to); j++) {
if (j % 2 == 1) {
cout << i << " ";
} else {
cout << i + j / 2 << " ";
}
}
}
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
test = int(input())
for _ in range(test):
n , l , r = [int(x) for x in input().split()]
start = 1
it = 1
if l == n*(n-1) + 1:
print('1')
continue
while start < l:
start += (n-it)*2
it += 1
if start != l:
it -= 1
start -= (n - it)*2
a = it
b = it+1
ok = True
while start < l:
if ok:
ok = False
else:
ok = True
b += 1
start += 1
# print(a , b , ok , '--------------------------')
while start <= r:
if b == n+1:
a += 1
b = a+1
if a == n:
# print('------------------')
print('1' , end=' ')
break
if ok:
print(a , end=' ')
ok = False
else:
print(b , end=' ')
b += 1
ok = True
# ok != ok
start += 1
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.math.*;
public class Main9
{
static class Pair
{
int x;
int y;
Pair(int x,int y)
{
this.x=x;
this.y=y;
}
}
static int mod=1000000007;
public static int[] sort(int[] a)
{
int n=a.length;
ArrayList<Integer> ar=new ArrayList<>();
for(int i=0;i<a.length;i++)
{
ar.add(a[i]);
}
Collections.sort(ar);
for(int i=0;i<n;i++)
{
a[i]=ar.get(i);
}
return a;
}
public static long pow(long a, long b)
{
long result=1;
while(b>0)
{
if (b % 2 != 0)
{
result=(result*a)%mod;
b--;
}
a=(a*a)%mod;
b /= 2;
}
return result;
}
public static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
public static long lcm(long a, long b)
{
return a*(b/gcd(a,b));
}
static ArrayList<ArrayList<Integer>> graph;
static public void main(String args[])throws IOException
{
int tt=i();
StringBuilder sb1=new StringBuilder();
for(int ttt=1;ttt<=tt;ttt++)
{
long n=l();
long l=l();
long r=l();
long edges=(n*((long)n-1))/2;
long sum=0;
long prev=0;
long index=-1;
for(long i=n;i>=1;i--)
{
sum=sum+(i-1)*2;
if(sum>=l)
{
index=i;
break;
}
prev=sum;
}
StringBuilder sb=new StringBuilder();
if(index==-1)
{
sb.append("1\n");
sb1.append(sb.toString());
}
else{
int flag=0;
long rem=n-index;
long starting=rem+1;
long fin_starting=starting;
long pp=0;
long count=0;
long extraL=l-prev;
if(extraL%2==0)
{
long num=extraL/2;
starting=(long)(starting+num);
sb.append(starting+" ");
count++;
}
else
{
starting=starting;
}
long ans=r-l+1;
long left=0;
long right=0;
if(fin_starting==starting)
{
long rr=extraL/2+1;
long ww=fin_starting+rr;
left=fin_starting;
right=(int)ww;
}
else
{
left=fin_starting;
right=starting+1;
}
right--;
long tttt=right;
long count1=count;
while(count<ans && flag==0)
{
while(left<n)
{
right=tttt+1;
while(right<=n && flag==0)
{
sb.append(left+" ");
count++;
if(count>=ans)
{
flag=1;
break;
}
sb.append(right+" ");
count++;
if(count>=ans)
{
flag=1;
break;
}
right++;
}
if(flag==1)
{
break;
}
left++;
tttt=left;
}
if(flag==1)
break;
if(count1==count)
break;
count1=count;
}
if(count<ans)
{
sb.append("1");
sb1.append(sb.toString()+"\n");
}
else
{
sb1.append(sb.toString()+"\n");
}}
}
System.out.print(sb1.toString());
}
/**/
static InputReader in=new InputReader(System.in);
static OutputWriter out=new OutputWriter(System.out);
public static long l()
{
String s=in.String();
return Long.parseLong(s);
}
public static void pln(String value)
{
System.out.println(value);
}
public static int i()
{
return in.Int();
}
public static String s()
{
return in.String();
}
}
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);
}
}
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();
}
}
class IOUtils {
public static int[] readIntArray(InputReader in, int size) {
int[] array = new int[size];
for (int i = 0; i < size; i++)
array[i] = in.Int();
return array;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Main {
static Parser parser = new Parser();
public static void main(String[] args) throws IOException {
int T = parser.parseInt();
for(int i = 0; i < T; i++){
solve();
}
}
static void solve() throws IOException{
int n = parser.parseInt();
long l = parser.parseLong();
long r = parser.parseLong();
int idx = 1;
long curr = 0;
while(curr + (n - idx) * 2 < l){
if(idx == n){
break;
}
curr += (n - idx) * 2;
idx += 1;
}
List<Integer> cycle = new ArrayList<>();
while(cycle.size() < r - curr + 1){
if(idx == n){
break;
}
for(int i = idx + 1; i <= n; i++){
cycle.add(idx);
cycle.add(i);
}
idx += 1;
}
cycle.add(1);
StringBuilder sb = new StringBuilder();
for(long i = l; i <= r; i++){
sb.append(cycle.get((int)(i - curr - 1)));
sb.append(' ');
}
System.out.println(sb.toString());
}
}
class Parser {
private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private static final Iterator<String> stringIterator = br.lines().iterator();
private static final Deque<String> inputs = new ArrayDeque<>();
void fill() throws IOException {
if(inputs.isEmpty()){
if(!stringIterator.hasNext()) throw new IOException();
inputs.addAll(Arrays.asList(stringIterator.next().split(" ")));
}
}
Integer parseInt() throws IOException {
fill();
if(!inputs.isEmpty()) {
return Integer.parseInt(inputs.pollFirst());
}
throw new IOException();
}
Long parseLong() throws IOException {
fill();
if(!inputs.isEmpty()) {
return Long.parseLong(inputs.pollFirst());
}
throw new IOException();
}
Double parseDouble() throws IOException {
fill();
if(!inputs.isEmpty()) {
return Double.parseDouble(inputs.pollFirst());
}
throw new IOException();
}
String parseString() throws IOException {
fill();
return inputs.pollFirst();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int t, n, s, lc;
long long l, r, k;
inline long long read() {
long long ans = 0;
char c = getchar();
while (c < 48 || c > 57) c = getchar();
while (c >= 48 && c <= 57)
ans = (ans << 3) + (ans << 1) + (c ^ 48), c = getchar();
return ans;
}
inline void write(int x) {
if (x > 9) write(x / 10);
putchar(x % 10 + 48);
}
int main() {
t = read();
while (t--) {
n = read(), l = read(), r = read(), k = 0;
for (register int i = 1; i < n; ++i) {
s = (n - i) << 1;
if (k + s < l) {
k += s;
continue;
}
if (k < l)
lc = l - k - 1, k = l - 1;
else
lc = 0;
for (register int j = lc; j < s - 1; ++j) {
++k;
if (k > r) break;
if (!(j & 1))
write(i);
else
write((j >> 1) + i + 1);
putchar(' ');
}
if (k > r) break;
++k;
if (k > r) break;
write(n), putchar(' ');
}
if (k < r) putchar(49);
putchar('\n');
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
ScanReader in = new ScanReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
DMinimumEulerCycle solver = new DMinimumEulerCycle();
solver.solve(1, in, out);
out.close();
}
static class DMinimumEulerCycle {
public void solve(int testNumber, ScanReader in, PrintWriter out) {
int t = in.scanInt();
loop:
while (t-- > 0) {
int n = in.scanInt();
long l = in.scanLong();
long r = in.scanLong();
long m = r - l + 1;
long last = (long) n * (n - 1) + 1;
if (m == 1 && l == r && r == last) {
out.println(1);
continue loop;
}
int arr[] = new int[n + 1];
long pre[] = new long[n + 1];
arr[1] = (n - 1) * 2;
for (int i = 2; i <= n; i++) arr[i] = arr[i - 1] - 2;
for (int i = 1; i <= n; i++) pre[i] = arr[i] + pre[i - 1];
int a = 0;
for (int i = 1; i <= n - 1; i++) if (pre[i] < l) a = i;
a++;
long next = l - pre[a - 1];
long b = a + ((next / 2) + (next % 2));
int turn = (int) (l % 2);
for (int i = 0; i < m; i++) {
if (i == m - 1 && r == last) {
out.print(1);
break;
}
if (turn == 1) {
out.print(a + " ");
} else {
out.print(b + " ");
b++;
if (b == (n + 1)) {
a++;
b = a + 1;
}
}
turn ^= 1;
}
out.println();
// 1 2 1 3 2 3
// 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5
}
}
}
static class ScanReader {
private byte[] buf = new byte[4 * 1024];
private int INDEX;
private BufferedInputStream in;
private int TOTAL;
public ScanReader(InputStream inputStream) {
in = new BufferedInputStream(inputStream);
}
private int scan() {
if (INDEX >= TOTAL) {
INDEX = 0;
try {
TOTAL = in.read(buf);
} catch (Exception e) {
e.printStackTrace();
}
if (TOTAL <= 0) return -1;
}
return buf[INDEX++];
}
public int scanInt() {
int I = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
I *= 10;
I += n - '0';
n = scan();
}
}
return neg * I;
}
private boolean isWhiteSpace(int n) {
if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
else return false;
}
public long scanLong() {
long I = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
I *= 10;
I += n - '0';
n = scan();
}
}
return neg * I;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
def main():
for _ in inputt():
n, l, r = inputi()
i = 1
l -= 1
if l == n * (n - 1):
print(1)
continue
while l >= 2 * (n - i):
l -= 2 * (n - i)
r -= 2 * (n - i)
i += 1
j = i + 1 + l // 2
while l < r:
if l % 2:
print(j, end = " ")
j += 1
if j > n:
i += 1
j = i + 1
elif i != n:
print(i, end = " ")
else:
print(1, end = " ")
l += 1
print()
# region M
# region fastio
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
for x in args:
file.write(str(x))
file.write(kwargs.pop("end", "\n"))
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
# region import
inputt = lambda t = 0: range(t) if t else range(int(input()))
inputi = lambda: map(int, input().split())
inputl = lambda: list(inputi())
from math import *
from heapq import *
from bisect import *
from itertools import *
from functools import reduce, lru_cache
from collections import Counter, defaultdict
import re, copy, operator, cmath
from builtins import *
# endregion
# region main
if __name__ == "__main__":
main()
# endregion
# endregion
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
long long t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> lol = {0};
for (long long i = n - 1; i >= 1; i--) {
lol.push_back(lol.back() + 2 * i);
}
for (long long x = l; x <= r; x++) {
if (x == n * (n - 1) + 1) {
cout << 1 << ' ';
} else {
long long block =
(long long)(lower_bound(lol.begin(), lol.end(), x) - lol.begin());
if (x % 2 == 1) {
cout << block << ' ';
} else {
long long h = x - lol[block - 1];
cout << h / 2 + block << ' ';
}
}
}
cout << '\n';
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int cas, z, x, w, h, a[500000];
long long n, m;
int main() {
scanf("%d", &cas);
while (cas--) {
scanf("%d%lld%lld", &z, &n, &m);
x = 1;
while (n > (z - x) * 2) {
if (x == z) {
m = 0;
printf("1");
break;
}
n -= (z - x) * 2;
m -= (z - x) * 2;
x++;
}
for (int i = 1; i <= z - x; i++) a[i * 2 - 1] = x, a[i * 2] = x + i;
if (m <= (z - x) * 2) {
for (int i = n; i <= m; i++) printf("%d ", a[i]);
} else {
for (int i = n; i <= (z - x) * 2; i++) printf("%d ", a[i]);
m -= (z - x) * 2;
x++;
while (m && (z != x)) {
for (int i = 1; i <= z - x; i++) a[i * 2 - 1] = x, a[i * 2] = x + i;
if (m <= (z - x) * 2) {
for (int i = 1; i <= m; i++) printf("%d ", a[i]);
m = 0;
} else {
for (int i = 1; i <= (z - x) * 2; i++) printf("%d ", a[i]);
m -= (z - x) * 2;
x++;
}
}
if (m) printf("1");
}
printf("\n");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class COVID {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
PrintWriter pw =new PrintWriter(System.out);
int t= sc.nextInt();
a: while(t-->0) {
int n= sc.nextInt();
long l=sc.nextLong();
long r= sc.nextLong();
long c=0;
int nc=1;
StringBuilder sb= new StringBuilder();
int i;
for( i=1;i<n;i++) {
if(2*(n-i)>=l)break;
l-=2*(n-i);
r-=2*(n-i);
}
all:for(;i<n;i++) {
for(int j=i+1;j<=n;j++) {
c++;
if(c>=l&&c<=r) {
sb.append(i+" ");
}
c++;
if(c>=l&&c<=r)
sb.append(j+" ");
if(c>r)break all;
}
}
c++;
if(c>=l&&c<=r)
sb.append("1 ");
l--;
pw.println(sb.toString());
}
pw.flush();
}
static long maximumSum(long[] a, long m) {
long sum=0;
long acc[] =new long[a.length];
for(int i=0;i<a.length;i++)
acc[i]=(int)((a[i]%m + ((i!=0)?acc[i-1]:0))%m);
TreeSet<Long> set= new TreeSet();
long min=m;
for(int i=0;i<acc.length;i++) {
set.add(acc[i]);
Long upper=set.ceiling(acc[i]+1);
if(upper==null) continue;
min=Math.min(min, upper-acc[i]);
}
return((m-min));
}
static class Pair implements Comparable<Pair>{
long x;int y,z;
public Pair(long a,int b, int c) {
this.x=a;y=b;this.z=c;
}
public int compareTo(Pair o) {
//return (this.x==o.x)?(this.y==o.y)?z-o.z:this.y-o.y:this.x-o.x;
return this.x>o.x?1:(this.x<o.x)?-1:0;
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(String s) throws FileNotFoundException{ br = new BufferedReader(new FileReader(s));}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine()," ,");
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++) {
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
if (sb.length() == 18) {
res += Long.parseLong(sb.toString()) / f;
sb = new StringBuilder("0");
}
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public boolean ready() throws IOException {
return br.ready();
}
}
public static void shuffle(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int r = i + (int) (Math.random() * (n - i));
int tmp = a[i];
a[i] = a[r];
a[r] = tmp;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
// package codeforces; /Users/attilaj/IdeaProjects/Codejam/src/codeforces/D1334.java
import java.util.Scanner;
public class D1334 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int tt = Integer.parseInt(sc.nextLine());
for (int t = 1; t <= tt; t++) {
String[] line = sc.nextLine().split(" ");
int n = Integer.parseInt(line[0]);
long l = Long.parseLong(line[1]);
long r = Long.parseLong(line[2]);
StringBuilder sb = new StringBuilder();
fill2(sb, n, 1, l - 1, r - 1);
System.out.println(sb.toString().substring(1));
}
}
private static boolean fill(StringBuilder sb, int n, int add, long l, long r) {
if (r < 0) {
return true;
} else if (n == 1) {
sb.append(' ').append(1);
return true;
} else if (l >= (n - 1) * 2) {
return fill(sb, n - 1, add + 1, l - (n - 1) * 2, r - (n - 1) * 2);
} else if (l < (n - 1) * 2) {
if (l % 2 == 1) {
sb.append(' ').append((l + 3) / 2 + add - 1);
l++;
}
long ll = l;
for (long i = l / 2; i <= r / 2 && i < n - 1; i++) {
sb.append(' ').append(add);
ll++;
if (i * 2 + 1 <= r) {
sb.append(' ').append(i + add + 1);
ll++;
}
}
l = ll;
return fill(sb, n - 1, add + 1, 0, r - l);
} else {
throw new IllegalStateException("??");
}
}
private static void fill2(StringBuilder sb, int n, int add, long l, long r) {
while (r >= 0) {
if (n == 1) {
sb.append(' ').append(1);
break;
} else if (l >= (n - 1) * 2) {
add++;
l -= (n - 1) * 2;
r -= (n - 1) * 2;
n--;
} else {
if (l % 2 == 1) {
sb.append(' ').append((l + 3) / 2 + add - 1);
l++;
}
long ll = l;
for (long i = l / 2; i <= r / 2 && i < n - 1; i++) {
sb.append(' ').append(add);
ll++;
if (i * 2 + 1 <= r) {
sb.append(' ').append(i + add + 1);
ll++;
}
}
n--;
add++;
r -= ll;
l = 0;
}
}
}
}
// n*(n-1)+1
// 2-->3
// 3-->7
// 4-->13
// 5-->21
// 1 2 1 3 2 3 1 4+2+1
// 1 2 1 3 1 4 2 3 2 4 3 4 1 6+4+2+1
// 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 1 8+6+4+2+1
// (n-1)*2 (n-2)*2
// 1 2
// 3 4
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace ::std;
const long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long add(long long a, long long b, long long m = MOD) {
if (a >= m) a %= m;
if (b >= m) b %= m;
if (a < 0) a += m;
if (b < 0) b += m;
long long res = a + b;
if (res >= m or res <= -m) res %= m;
if (res < 0) res += m;
return res;
}
long long mul(long long a, long long b, long long m = MOD) {
if (a >= m) a %= m;
if (b >= m) b %= m;
if (a < 0) a += m;
if (b < 0) b += m;
long long res = a * b;
if (res >= m or res <= -m) res %= m;
if (res < 0) res += m;
return res;
}
long long pow_mod(long long a, long long b, long long m = MOD) {
long long res = 1LL;
a = a % m;
while (b) {
if (b & 1) res = mul(res, a, m);
b >>= 1;
a = mul(a, a, m);
}
return res;
}
long long fastexp(long long a, long long b) {
long long res = 1LL;
while (b) {
if (b & 1) res = res * a;
b >>= 1;
a *= a;
}
return res;
}
int gcdExtendido(int a, int b, int *x, int *y) {
if (a == 0) {
*x = 0;
*y = 1;
return b;
}
int x1, y1;
int gcd = gcdExtendido(b % a, a, &x1, &y1);
*x = y1 - (b / a) * x1;
*y = x1;
return gcd;
}
int modInverso(int a, int m) {
int x, y;
int g = gcdExtendido(a, m, &x, &y);
if (g != 1)
return -1;
else
return (x % m + m) % m;
}
const int N = 100000 + 5;
int n;
long long l, r;
int ans[N];
long long f(int x) { return 2LL * n * x - 1LL * x * (x + 1); }
int main() {
int t;
scanf("%d", &(t));
while (t--) {
scanf("%d", &(n));
scanf("%lld %lld", &(l), &(r));
int len = r - l + 1;
if (r == 1LL * n * (n - 1) + 1) {
ans[r - l] = 1;
r -= 1;
}
int lo = 1, hi = n;
while (lo < hi) {
int mi = lo + (hi - lo) / 2;
if (f(mi) < l)
lo = mi + 1;
else
hi = mi;
}
l -= f(lo - 1);
r -= f(lo - 1);
int pos = 0;
int x = lo;
int y = lo + (l + 1) / 2;
bool goX = l & 1;
for (int i = l; i <= r; i++) {
if (goX) {
ans[pos++] = x;
} else {
ans[pos++] = y;
y += 1;
}
if (y == n + 1) {
x += 1;
y = x + 1;
}
goX = !goX;
}
for (int i = 0; i < len; i++) {
printf("%d%c", ans[i], " \n"[i + 1 == len]);
}
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
cin >> t;
long long cur = 0;
while (t--) {
cur = 0;
long long n, l, r;
cin >> n >> l >> r;
for (int i = 1; i <= n; i++) {
if (cur + 2 * (n - i) < l)
cur += 2 * (n - i);
else {
while (l <= r && i <= n && l <= cur + 2 * (n - i)) {
if (l % 2)
cout << i << " ";
else
cout << (l - cur) / 2 + i << " ";
l++;
}
cur += 2 * (n - i);
}
}
if (r - l == 0) cout << "1";
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int T;
int n;
long long l, r;
void out(long long sum, int now, int num, long long i) {
if (now == n) {
if (i != l) printf(" ");
printf("1");
return;
}
for (i; i <= min(sum + num - 1, r); i++) {
int N;
if ((i - sum + 1) % 2)
N = now;
else
N = now + (i - sum + 1) / 2;
if (i == l) {
printf("%d", N);
} else
printf(" %d", N);
}
}
int main() {
scanf("%d", &T);
while (T--) {
long long sum = 1;
scanf("%d%lld%lld", &n, &l, &r);
long long now = 1, num = (n - now) * 2;
for (now = 1; now <= n; now++) {
num = num = (n - now) * 2;
num = max(num, 1 * 1ll);
if (sum <= l && l <= sum + num) {
out(sum, now, num, l);
} else if (l <= sum && sum + num <= r) {
out(sum, now, num, sum);
} else if (sum <= r && r <= sum + num) {
out(sum, now, num, sum);
break;
}
sum += num;
}
printf("\n");
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long int n, l, r;
cin >> n >> l >> r;
long long int sum = 0;
long long int start = 1;
int done = 0;
while (sum < l) {
sum += 2 * (n - start);
start++;
if (start > n) {
cout << 1 << " " << endl;
done = 1;
break;
}
}
if (done) continue;
start--;
sum -= 2 * (n - start);
long long int next = start + 1;
long long int cur = 0;
for (long long int i = sum + 1; i <= r; i++) {
if (i < l) {
if (i % 2 == 0) next++;
} else {
if (i % 2 == 1)
cout << start << " ";
else {
cout << next << " ";
next++;
}
}
cur++;
if (cur == 2 * (n - start)) {
cur = 0;
start = start + 1;
next = start + 1;
if (start == n) start = 1;
}
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
size_t T;
cin >> T;
while (T--) {
long long int n, l, r;
cin >> n >> l >> r;
long long int t = 1;
long long int k = 1;
while (k < l && t != n) {
k += 2 * (n - t++);
}
if (k > l) {
k -= 2 * (n - --t);
}
if (t == n) {
cout << 1 << endl;
continue;
}
long long int difference = l - k;
long long int c = difference / 2 + t + 1;
if (difference % 2 == 0) {
long long int current = l;
while (current <= r) {
cout << t << " ";
++current;
if (current <= r) {
cout << c++ << " ";
++current;
if (c > n) {
c = ++t + 1;
if (t == n) {
if (current <= r) cout << 1;
break;
}
}
}
}
} else {
cout << c << " ";
++l;
difference = l - k;
c = difference / 2 + t + 1;
if (c > n) {
c = ++t + 1;
if (t == n) {
cout << 1 << endl;
break;
}
}
long long int current = l;
while (current <= r) {
cout << t << " ";
++current;
if (current <= r) {
cout << c++ << " ";
++current;
if (c > n) {
++t;
c = t + 1;
if (t == n) {
if (current <= r) cout << 1;
break;
}
}
}
}
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
void solve(int qq) {
long long n;
cin >> n;
long long l, r;
cin >> l >> r;
long long cnt = 0;
long long odd = 0;
long long start = 0;
if (l == n * (n - 1) + 1) {
cout << 1 << endl;
return;
}
for (long long i = 1; i < n; i++) {
cnt += (n - i) * 2;
if (l <= cnt) {
odd = i;
start = cnt - (n - i) * 2 + 1;
break;
}
}
long long even = odd + ((l - start) / 2 + 1);
for (long long i = l; i <= r; i++) {
if (i % 2 == 0) {
cout << even << " ";
even++;
if (even > n) {
odd++;
even = odd + 1;
if (odd == n) {
odd = 1;
}
}
} else {
cout << odd << " ";
}
}
cout << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin >> t;
for (int i = 1; i <= t; i++) {
solve(i);
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
for (int tc = 0; tc < (t); tc += 1) {
long long n, l, r;
cin >> n >> l >> r;
l--;
bool last1 = false;
if (r == n * (n - 1) + 1) {
last1 = true;
r--;
if (r == l) {
cout << "1\n";
continue;
}
}
long long i = 1;
long long ptr = 0;
while (ptr + (n - i) * 2 <= l) {
ptr += (n - i) * 2;
i++;
}
long long j = i;
j += (l - ptr) / 2;
if (l % 2 == 1) {
cout << j + 1 << " ";
j++;
l++;
}
if (j != n) {
j++;
} else {
i++;
j = i + 1;
}
while (r > l) {
if (r - l == 1) {
cout << i << " ";
break;
}
cout << i << " " << j << " ";
if (j != n) {
j++;
} else {
i++;
j = i + 1;
}
r -= 2;
}
if (last1) cout << "1";
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long T, t, n, N, l, L, R;
vector<int> v;
pair<int, int> calcL() {
for (n = l = 0; n < N; n++) {
if ((l + (N - (n + 1))) > (L / 2)) break;
l += (N - (n + 1));
}
return pair<int, int>(n, 1 + n + (int)((L / 2) - l));
}
void fill() {
v.clear();
pair<int, int> curr = calcL();
for (n = L; n <= R; n++) {
if (curr.first >= (N - 1)) {
curr.first = 0;
}
if (curr.second >= N) {
curr.second = 0;
}
if (n & 1) {
v.push_back(curr.second);
if (curr.second == (N - 1)) {
curr = pair<int, int>((curr.first + 1), (curr.first + 2));
} else
curr.second++;
} else
v.push_back(curr.first);
}
}
int main() {
cin >> T;
for (t = 0; t < T; t++) {
cin >> N >> L >> R;
L--;
R--;
fill();
for (n = 0; n <= (R - L); n++) {
printf("%d ", 1 + v[n]);
}
printf("\n");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def solve(n,l,r):
fir,st = 0,1
while st < n:
x = 2*(n-st)
if fir+x >= l:
break
fir += x
st += 1
if st == n:
return [1]
ans = []
for z in range(st+1,n+1):
ans.append(st)
ans.append(z)
st += 1
while len(ans) < r-fir:
if st == n:
ans.append(1)
else:
for z in range(st+1,n+1):
ans.append(st)
ans.append(z)
st += 1
return ans[l-fir-1:r-fir]
def main():
for _ in range(int(input())):
n,l,r = map(int,input().split())
print(*solve(n,l,r))
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long n, l, r;
int Go() {
scanf("%lld %lld %lld", &n, &l, &r);
long long len = r - l + 1;
long long x = n;
long long y = 0;
long long start = 0;
while (1) {
y += 2 * (x - 1);
x--;
start++;
if (x <= 0 || y >= l) break;
}
long long k = y - 2 * x;
long long d = (l - k + 1) / 2;
long long add = start + d;
vector<long long> Ans;
long long count = 0;
if (l % 2 == 0) {
Ans.push_back(add++);
count++;
}
while (count < len) {
if (add == n + 1) {
start++;
add = start + 1;
}
Ans.push_back(start);
count++;
if (count == len) break;
Ans.push_back(add);
count++;
add++;
}
if (r == n * (n - 1) + 1) {
Ans.pop_back();
Ans.push_back(1);
}
for (int i = 0; i < Ans.size(); ++i) printf("%lld ", Ans[i]);
puts("");
return 0;
}
int main() {
int t;
if (1)
scanf("%d", &t);
else
t = 1;
while (t--) Go();
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import sys
import math
input = sys.stdin.readline
from functools import cmp_to_key;
def pi():
return(int(input()))
def pl():
return(int(input(), 16))
def ti():
return(list(map(int,input().split())))
def ts():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
mod = 1000000007;
f = [];
def fact(n,m):
global f;
f = [1 for i in range(n+1)];
f[0] = 1;
for i in range(1,n+1):
f[i] = (f[i-1]*i)%m;
def fast_mod_exp(a,b,m):
res = 1;
while b > 0:
if b & 1:
res = (res*a)%m;
a = (a*a)%m;
b = b >> 1;
return res;
def inverseMod(n,m):
return fast_mod_exp(n,m-2,m);
def ncr(n,r,m):
if r == 0: return 1;
return ((f[n]*inverseMod(f[n-r],m))%m*inverseMod(f[r],m))%m;
def main():
D();
def xdfs(root, v, sub, parent):
st = [root];
while len(st) > 0:
node = st.pop();
for i in range(len(v[node])):
if v[node][i] != node:
st.append(v[node][i]);
def X():
try:
t = pi();
while t:
t -= 1;
n = pi();
v = [[] for i in range(n)];
for i in range(n-1):
[x,y] = ti();
v[x-1].append(y-1);
v[y-1].append(x-1);
m = pi();
p = ti();
e = [0 for i in range(n-1)];
sub = [0 for i in range(n)];
dfs(0,v,sub,-1)
for i in range(1,n):
e[i-1] = (sub[i]*(n-sub[i]));
if len(p) < n-1:
while len(p) < n-1:
p.append(1);
p = sorted(p);
if len(p) > n-1:
x = 1;
for i in range(n-2,len(p)):
x = (x*p[i]);
while len(p) > n-2:
p.pop();
p.append(x);
e = sorted(e);
res = 0;
for i in range(n-1):
res = (res+(p[i]*e[i]))%mod;
print(res);
except:
print(sys.exc_info());
def bfs(v,root):
q = [root,None];
l = 0;
visited = [0 for i in range(len(v))];
dist = [0 for i in range(len(v))];
while len(q) > 0:
node = q.pop(0);
if node is not None:
visited[node] = 1;
dist[node] = l;
for i in range(len(v[node])):
if visited[v[node][i]] == 0:
visited[v[node][i]] = 1;
q.append(v[node][i]);
else:
l += 1;
if len(q) != 0:
q.append(None);
return dist;
def B():
n = pi();
v = [[] for i in range(n)];
for i in range(n-1):
[x,y] = ti();
v[x-1].append(y-1);
v[y-1].append(x-1);
leafs = [];
for i in range(n):
if len(v[i]) == 1:
leafs.append(i);
mn = 1;
d = bfs(v,leafs[0]);
for i in range(1,len(leafs)):
if d[leafs[i]] % 2 != 0:
mn = 3;
break;
count = 0;
for i in range(n):
f = 0;
for j in range(len(v[i])):
if len(v[v[i][j]]) == 1:
f = 1;
break;
if f: count += 1;
mx = n-1-len(leafs)+count;
print(mn,mx)
def D():
t = pi();
while t:
t -= 1;
[n,l,r] = ti();
p = 0;
i = 1;
nxt = 0;
res = [];
st = 0;
while i <= n and p <= l:
if p+2*(n-i) >= l:
d = l-p-1;
if d%2 == 0:
st = i;
nxt = i+1+(d/2);
if d%2 != 0:
st = i+1+((d-1)/2);
p = l;
break;
p += 2*(n-i);
i += 1;
while i <= n and p <= r:
if nxt != 0:
while p <= r and nxt <= n:
if p <= r:
res.append(i);
p += 1;
if p <= r:
res.append(int(nxt));
p += 1;
nxt += 1;
i += 1;
nxt = i+1;
else:
nxt = st+1;
# if nxt == n:
# i += 1;
# nxt = i+1;
res.append(int(st));
p += 1;
while p <= r and nxt <= n:
if p <= r:
res.append(i);
p += 1;
if p <= r:
res.append(int(nxt));
p += 1;
nxt += 1;
i += 1;
nxt = i+1;
if r == (n-1)*(n-1)+n: res.append(1);
print(*res, sep=" ");
main();
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
import random
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
from math import factorial as f
# def ncr(x, y):
# return f(x) // (f(y) * f(x - y))
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def solve(n):
return (n*(n+1))//2
t = int(input())
for _ in range(t):
n,l,r = map(int,input().split())
# print((2*n - 1)**2,4*l)
if 2*(n)*n - (n)*(n+1) + 1 == l:
print(1)
continue
# print(2*(n)*n - (n)*(n+1))
z_1 = (2*n - 1 + sqrt((2*n - 1)**2 - 4*l))
z_2 = (2*n - 1 - sqrt((2*n - 1)**2 - 4*l))
z1 = ceil(min(z_1,z_2) / 2)
# print(2*(n)*n - (n)*(n+1))
# z_1 = 2*n - 1 + sqrt((2*n - 1)**2 + 4*r)
# z_2 = 2*n - 1 - sqrt((2*n - 1)**2 + 4*r)
# z2 = max(-z_1,-z_2) // 2
# print(z1)
# print(z1)
z0 = z1-1
la = l
l-= 2*(z0)*n - (z0)*(z0+1)
# print(z1,l)l
# z0 = z23
# 2 1 3
# 3 3 6
# 99995 9998900031 9998900031
# r-= 2*(z0)*n - (z0)*(z0+1)
# z1+=1
# print(z1)
# if z1 == n:
# print(1)
# continue
# z2+=1
ans = []
if l%2==0:
k1 = l//2
l = la
else:
ans.append(int(z1))
k1 = (l+1)//2
l = la
if r-l +1 == len(ans):
print(*ans)
continue
l+=1
# print(k1)
ha = z1+1 + (k1-1)
ans.append(int(ha))
l+=1
cnt = max(ans)+1
while l!=r+1:
if ans[-1] == n:
z1+=1
cnt = z1+1
if z1 == n+1 or z1 == n:
ans.append(1)
break
else:
ans.append(int(z1))
else:
if l%2 != 0:
ans.append(int(z1))
else:
ans.append(int(cnt))
cnt+=1
l+=1
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import sys
input = sys.stdin.readline
T=int(input())
for tests in range(T):
n,l,r=map(int,input().split())
begin=1
while l>(n-begin)*2+1:
if begin==n:
break
l-=(n-begin)*2
r-=(n-begin)*2
begin+=1
#print(begin,l,r)
if begin==n:
ANS=[n,1]
else:
ANS=[]
while len(ANS)<=r:
if begin==n:
ANS.append(1)
break
for j in range(begin+1,n+1):
ANS.append(begin)
ANS.append(j)
begin+=1
#print(ANS)
sys.stdout.write(" ".join(map(str,ANS[l-1:r]))+"\n")
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
from sys import stdin, gettrace
from math import sqrt
if not gettrace():
def input():
return next(stdin)[:-1]
# def input():
# return stdin.buffer.readline()
def main():
def solve():
n,l,r = map(int, input().split())
lv = int((2*n+1 - sqrt((2*n-1)**2 -4*(l-1)))/2)
lvs = -2*n+2*n*lv-lv*lv+lv
lrd = l - lvs - 1
res = []
i = lv
j = lv+lrd//2 + 1
if l%2 == 0:
res = [j]
if j < n:
j+=1
else:
i+=1
j = i+1
for _ in range(l-1, r, 2):
res += [i,j]
if j < n:
j += 1
else:
i +=1
j = i+1
if r == n*(n-1)+1:
res[r-l] = 1
print(*res[:r-l+1])
q = int(input())
for _ in range(q):
solve()
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class D {
Reader source;
BufferedReader br;
StringTokenizer in;
PrintWriter out;
public String nextToken() throws Exception {
while (in == null || !in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public int nextInt() throws Exception {
return Integer.parseInt(nextToken());
}
public long nextLong() throws Exception {
return Long.parseLong(nextToken());
}
public double nextDouble() throws Exception {
return Double.parseDouble(nextToken());
}
public void solve() throws Exception {
int t = nextInt();
for (int i = 0; i < t; i++) {
int n = nextInt();
long l = nextLong();
long r = nextLong();
int a = 1;
while (a < n && l > 2 * (n - a)) {
l = l - 2 * (n - a);
r = r - 2 * (n - a);
a++;
}
if (a == n) a = 1;
int b = a + 1;
while (l > 2) {
l = l - 2;
r = r - 2;
b++;
}
//out.println("a=" + a + " b=" + b + " l=" + l + " r=" + r);
while (l <= r) {
if (l % 2 == 1) {
out.print(a + " ");
} else {
out.print(b + " ");
b++;
if (b > n) {
a++;
if (a == n) a = 1;
b = a + 1;
}
}
l++;
}
out.println();
}
}
public void run() throws Exception {
source = OJ ? new InputStreamReader(System.in) : new FileReader("D.in");
br = new BufferedReader(source);
out = new PrintWriter(System.out);
solve();
out.flush();
}
public static void main(String[] args) throws Exception {
new D().run();
}
private boolean OJ = System.getProperty("ONLINE_JUDGE") != null;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from heapq import heapify,heappush,heappop
def main():
for _ in range(int(input())):
n,l,r=map(int,input().split())
l-=1
ptrn=[]
for i in range(2,10**5+2):
ptrn.append(1)
ptrn.append(i)
# print(ptrn[:10])
lth=2*(n-1)
ans=[]
sf=0
ps=0
while lth>0:
# print(lth,sf,lth,max(ps,l),min(r,ps+lth))
for i in range(max(ps,l),min(r,ps+lth)):
ans.append(ptrn[i-ps]+sf)
ps+=lth
lth-=2
sf+=1
if r==n*(n-1)+1:
ans.append(1)
print(*ans)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
for i in range(int(input())):
n, l, r = map(int, input().split())
l -= 1
r -= 1
c = 0
for j in range(n):
if (c + (n - j - 1) * 2 < l or c > r):
c += (n - j - 1) * 2
continue
for k in range(j + 1, n):
if (l <= c <= r):
print(j + 1, end = ' ')
c += 1
if (l <= c <= r):
print(k + 1, end = ' ')
c += 1
if (l <= c <= r):
print(1)
else:
print('')
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python2
|
rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())
def solve(N, L, R):
L -= 1; R -= 1
if N == 1:
return [1][L:R+1]
if N == 2:
return [1,2,1][L:R+1]
if N == 3:
return [1,2,1,3,2,3,1][L:R+1]
page = 2*N - 2
k = 1
left = 0
while left + page <= L:
left += page
k += 1
page -= 2
if page == 0: page = 1
def write(k):
for j in xrange(k+1, N+1):
bns.append(k)
bns.append(j)
if k == N:
bns.append(1)
bns = []
while len(bns) < R - left + 1:
write(k)
k += 1
return bns[L-left:R-left+1]
for tc in xrange(rri()):
ans = solve(*rrm())
print " ".join(map(str, ans))
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.ArrayList;
import java.util.List;
public class CE85D {
public static void main(String[] args) throws NumberFormatException, IOException {
FastReader sc=new FastReader();
int t = sc.nextInt();
for(int x=0; x<t; x++) {
long n = sc.nextLong();
long l = sc.nextLong();
long r = sc.nextLong();
long total = (n)*(n-1)+1;
//System.out.println("totes" + total);
// if(l == total && r==total) {
// System.out.println(1);
// continue;
// }
long first = 1;
long second = 0;
long sum = 0;
while(true) {
long next = 2*(n-first);
//System.out.println(next);
if(sum + next < l) {
first += 1;
sum += next;
} else {
break;
}
if(next == 2) {
break;
}
}
//System.out.println("sumdy" + sum);
long diff = l - sum;
second = first + 1 + ((diff-1)/2);
//System.out.println("fgege" + first + " " + second);
while(l <= r) {
//System.out.println("ell"+ l);
if(l == total) {
System.out.print("1 ");
} else {
if(l%2 == 0) { //second
System.out.print(second + " ");
if(second < n) {
second += 1;
} else {
first += 1;
second = first + 1;
}
} else {
System.out.print(first + " ");
}
}
l++;
}
System.out.println();
}
}
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 n = sc.nextInt();
for(int y=0; y<n; y++){
int x = sc.nextInt();
}
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int d = sc.nextInt();
int e = sc.nextInt();
*/
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
/*
*/
public class A {
static FastReader sc=null;
static int nax=(int)1e6 +10;
public static void main(String[] args) {
sc=new FastReader();
PrintWriter out=new PrintWriter(System.out);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long l=sc.nextLong(),r=sc.nextLong();
long l1=l;
int curr=1;
while(l>2*(n-curr) && curr<=n) {
l-=2*(n-curr);
curr++;
//System.out.println(l+" ");
}
//CURR indicates which starting point we get to and now
int from=(int)(l/2+curr);
int ans[]=new int[nax];
//System.out.println(l+" "+r+" "+from);
Arrays.fill(ans, -1);
int j=0;
if(l%2==0) {
ans[0]=from;
from++;
j++;
}
else {
from++;
}
while(from<=n) {
ans[j++]=curr;
ans[j++]=from++;
}
curr++;
int len=(int)(r-l1+1)+5;
while(j<len && curr<n) {
for(int k=(curr+1);k<=n && j<len;k++) {
ans[j++]=curr;
ans[j++]=k;
}
curr++;
}
if(j<nax)ans[j++]=1;
//System.out.println(len);
for(int i=0;i<len-5;i++) {
//if(ans[i]==-1)break;
out.print(ans[i]+" ");
}
out.println();
}
out.close();
}
static void reverseSort(int a[]) {
ArrayList<Integer> al=new ArrayList<>();
for(int i:a)al.add(i);
Collections.sort(al,Collections.reverseOrder());
for(int i=0;i<a.length;i++)a[i]=al.get(i);
}
static int gcd(int a,int b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static long gcd(long a,long b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static void reverse(int a[]) {
int n=a.length;
int b[]=new int[n];
for(int i=0;i<n;i++)b[i]=a[n-1-i];
for(int i=0;i<n;i++)a[i]=b[i];
}
static void 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);
}
static void print(int a[]) {
for(int e:a) {
System.out.print(e+" ");
}
System.out.println();
}
static void print(long a[]) {
for(long e:a) {
System.out.print(e+" ");
}
System.out.println();
}
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[] readArray(int n) {
int a[]=new int [n];
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
}
return a;
}
long[] readArrayL(int n) {
long a[]=new long [n];
for(int i=0;i<n;i++) {
a[i]=sc.nextLong();
}
return a;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Calendar;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.SortedSet;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
/**
* #
*
* @author pttrung
*/
public class D_Edu_Round_73 {
public static long MOD = 1000000007;
public static void main(String[] args) throws FileNotFoundException {
// PrintWriter out = new PrintWriter(new FileOutputStream(new File(
// "output.txt")));
PrintWriter out = new PrintWriter(System.out);
Scanner in = new Scanner();
int T = in.nextInt();
for (int z = 0; z < T; z++) {
int n = in.nextInt();
long l = in.nextLong();
long r = in.nextLong();
for (long i = l; i <= r; i++) {
int st = 1;
int ed = n;
int result = 0;
while (st <= ed) {
int mid = (st + ed) / 2;
if (cal(mid, n) < i) {
result = mid;
st = mid + 1;
} else {
ed = mid - 1;
}
}
//System.out.println(i + " " + result + " " + cal(result, n));
if (result == n - 1) {
out.print(1);
} else {
long index = i - cal(result, n);
if (index % 2 != 0) {
out.print((result + 1) + " ");
} else {
out.print((result + 1 + (index / 2)) + " ");
}
}
}
out.println();
}
out.close();
}
static long cal(long index, long n) {
if (index == n) {
return 1 + cal(n - 1, n);
}
long result = 2 * n * index - (index * (index + 1));
return result;
}
public static int[] KMP(String val) {
int i = 0;
int j = -1;
int[] result = new int[val.length() + 1];
result[0] = -1;
while (i < val.length()) {
while (j >= 0 && val.charAt(j) != val.charAt(i)) {
j = result[j];
}
j++;
i++;
result[i] = j;
}
return result;
}
public static boolean nextPer(int[] data) {
int i = data.length - 1;
while (i > 0 && data[i] < data[i - 1]) {
i--;
}
if (i == 0) {
return false;
}
int j = data.length - 1;
while (data[j] < data[i - 1]) {
j--;
}
int temp = data[i - 1];
data[i - 1] = data[j];
data[j] = temp;
Arrays.sort(data, i, data.length);
return true;
}
public static int digit(long n) {
int result = 0;
while (n > 0) {
n /= 10;
result++;
}
return result;
}
public static double dist(long a, long b, long x, long y) {
double val = (b - a) * (b - a) + (x - y) * (x - y);
val = Math.sqrt(val);
double other = x * x + a * a;
other = Math.sqrt(other);
return val + other;
}
public static class Point implements Comparable<Point> {
int x, y;
public Point(int start, int end) {
this.x = start;
this.y = end;
}
public String toString() {
return x + ":" + y;
}
@Override
public int hashCode() {
int hash = 5;
hash = 47 * hash + this.x;
hash = 47 * hash + this.y;
return hash;
}
@Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final Point other = (Point) obj;
if (this.x != other.x) {
return false;
}
if (this.y != other.y) {
return false;
}
return true;
}
@Override
public int compareTo(Point o) {
if (x != o.x) {
return Integer.compare(x, o.x);
}
return Integer.compare(y, o.y);
}
}
public static class FT {
long[] data;
FT(int n) {
data = new long[n];
}
public void update(int index, long value) {
while (index < data.length) {
data[index] += value;
index += (index & (-index));
}
}
public long get(int index) {
long result = 0;
while (index > 0) {
result += data[index];
index -= (index & (-index));
}
return result;
}
}
public static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
public static long pow(long a, int b) {
if (b == 0) {
return 1;
}
if (b == 1) {
return a;
}
long val = pow(a, b / 2);
if (b % 2 == 0) {
return val * val;
} else {
return val * (val * a);
}
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() throws FileNotFoundException {
// System.setOut(new PrintStream(new BufferedOutputStream(System.out), true));
br = new BufferedReader(new InputStreamReader(System.in));
// br = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt"))));
}
public String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
throw new RuntimeException();
}
}
return st.nextToken();
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
st = null;
try {
return br.readLine();
} catch (Exception e) {
throw new RuntimeException();
}
}
public boolean endLine() {
try {
String next = br.readLine();
while (next != null && next.trim().isEmpty()) {
next = br.readLine();
}
if (next == null) {
return true;
}
st = new StringTokenizer(next);
return st.hasMoreTokens();
} catch (Exception e) {
throw new RuntimeException();
}
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.StringTokenizer;
public class D {
public static void main(String[] args) throws Exception
{
new D().run();
}
LinkedList<Integer> inversionLocations = new LinkedList<Integer>();
int K;
int inversions;
char[] chars;
ArrayList<Integer> toAdd = new ArrayList<Integer>();
public void run() throws Exception
{
BufferedReader file = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(file.readLine());
int T = Integer.parseInt(st.nextToken());
PrintWriter pout = new PrintWriter(System.out);
for(int z = 0;z<T;z++)
{
st = new StringTokenizer(file.readLine());
long K = Long.parseLong(st.nextToken());
long L = Long.parseLong(st.nextToken());
long R = Long.parseLong(st.nextToken());
StringBuilder sb = new StringBuilder("");
for(long i = L;i<=R;i++)
{
sb.append(getIndex(K,i-1));
sb.append(" ");
}
pout.println(sb.toString());
}
pout.flush();
}
public long getIndex(long K, long index)
{
if(index == K*(K-1))
return 1;
return getIndexHelper(K, index);
}
public long blockSum(long K, long x)
{
if(x == 0)
return 0;
long first = (K-1)*2;
long last = (K-x)*2;
long terms = ((first - last)/2)+1;
return (first + last) * terms/2;
}
public int getBlock(long K, long index)
{
long L = 0;
long R = K+1;
long M = (R+L)/2;
long ceil = K+1;
while(R-L > 1)
{
M = ((R+L)/2);
long sum = blockSum(K, M);
//System.out.println(L+" "+R+" "+M);
if(sum >= index)
{
ceil = M;
R = M;
}else {
L = M;
}
}
return (int)(ceil);
}
public long getIndexHelper(long K, long index)
{
int block = getBlock(K, index+1)-1;
//System.out.println(index+"_"+block);
//System.out.println("K = "+K+", index "+index+" block is "+block);
long remaining = index - blockSum(K, block);
//System.out.println("remaining: " + remaining);
if((remaining & 1) == 0)
return block+1;
return (block+1) + (remaining/2 + 1);
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
from sys import stdin
def allWays(start, verts, done, stack, n):
global valid
if not valid:
return
stack.append(start)
if len(done) == len(verts):
print(stack)
stack.pop()
valid = False
return
for x in range(1,n+1):
if start != x and not (start,x) in done:
done.add((start,x))
allWays(x,verts,done,stack,n)
done.remove((start,x))
stack.pop()
'''
for y in range(1,10):
verts = set([((x//y) + 1, (x%y)+1) for x in range(y**2)])
for x in range(1,y+1):
verts.remove((x,x))
valid = True
print(y, end=' ')
allWays(1, verts, set(), [], y)
'''
def order(n,x):
out = []
for y in range(x+1,n+1):
out.append(x)
out.append(y)
return out
for case in range(int(stdin.readline())):
n,l,r = [int(x) for x in stdin.readline().split()]
end1 = False
if r == n*(n-1) + 1:
end1 = True
r -= 1
if l == n*(n-1) + 1:
print(1)
else:
x = 1
while l > 2*(n-x):
l -= 2*(n-x)
r -= 2*(n-x)
x += 1
out = order(n,x)
r -= 2*(n-x)
x += 1
while r > 0:
out += order(n,x)
r -= 2*(n-x)
x += 1
if end1:
out += [1]
if r != 0:
realOut = out[l-1:r]
else:
realOut = out[l-1:]
print(' '.join([str(b) for b in realOut]))
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long n, l, r;
cin >> n >> l >> r;
if (n == 2) {
for (long long i = l - 1; i < r; i++) {
if (i & 1)
cout << 2 << " ";
else
cout << 1 << " ";
}
return;
}
if ((r - 1) / 2 + 1 < n) {
vector<long long> ans;
long long vl = l - 1;
vl /= 2;
l -= vl * 2;
r -= vl * 2;
vl += 2;
while (ans.size() < r - l + 1) {
ans.push_back(1);
ans.push_back(vl);
vl++;
}
for (long long i = l - 1; i < r; i++) {
cout << ans[i] << " ";
}
return;
}
long long lb = -1, rb = n - 1;
while (rb - lb != 1) {
long long mid = lb + rb >> 1;
if ((n - 1) * 2 + (n - 2) * (n - 1) - (n - 1 - mid) * (n - 2 - mid) >= l)
rb = mid;
else
lb = mid;
}
long long from = rb;
lb = -1, rb = n - 1;
while (rb - lb != 1) {
long long mid = rb + lb >> 1;
if ((n - 1) * 2 + (n - 2) * (n - 1) - (n - 1 - mid) * (n - 2 - mid) >= r)
rb = mid;
else
lb = mid;
}
vector<long long> ans;
long long to = rb;
long long rto = from;
if (from == 0) {
for (long long i = 2; i <= n; i++) {
ans.push_back(1);
ans.push_back(i);
}
from++;
}
while (from <= min(to, n - 2)) {
ans.push_back(from + 1);
for (long long i = from + 2; i < n; i++) {
ans.push_back(i);
ans.push_back(from + 1);
}
ans.push_back(n);
from++;
}
ans.push_back(1);
rto--;
long long val = 0;
if (rto != -1)
val = (n - 1) * 2 + (n - 2) * (n - 1) - (n - 1 - rto) * (n - 2 - rto);
r -= val;
l -= val;
for (long long i = l - 1; i < r; i++) cout << ans[i] << " ";
}
signed main() {
long long t;
cin >> t;
while (t--) {
solve();
cout << endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10;
long long n;
long long pre[N];
long long cal(long long x) {
if (x == n * (n - 1) + 1) return 1;
long long p = lower_bound(pre + 1, pre + n + 1, x) - pre;
long long b = x - pre[p - 1];
if (b & 1)
return p;
else
return p + b / 2;
}
signed main() {
long long t;
cin >> t;
while (t--) {
long long l, r;
cin >> n >> l >> r;
for (long long i = 1; i <= n; i++) pre[i] = pre[i - 1] + 2 * (n - i);
for (long long i = l; i <= r; i++) cout << cal(i) << " ";
puts("");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n;
ll l, r;
cin >> n >> l >> r;
ll pre = 1;
VI ans;
for (int s = 1; s < n; s++) {
ll len = 2 * (n - s);
if (l < pre + len && r >= pre) {
for (int i = s + 1; i <= n; i++) {
if (pre >= l && pre <= r) {
ans.push_back(s);
}
pre++;
if (pre >= l && pre <= r) {
ans.push_back(i);
}
pre++;
}
} else {
pre += len;
}
}
if (r == n * 1ll * (n - 1) + 1) ans.push_back(1);
for (int u : ans) cout << u << ' ';
cout << '\n';
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000000007;
inline long long int gcd(long long int a, long long int b) {
return (b == 0) ? a : gcd(b, a % b);
}
inline long long int lcm(long long int a, long long int b) {
return (a * b) / gcd(a, b);
}
inline long long int mymod(long long int A, long long int M) {
return ((A % M) + M) % M;
}
template <class type>
type power(type x, long long int n) {
type temp;
long long int y = n;
if (y == 0) return 1;
temp = power(x, y / 2);
return ((y % 2) ? ((y > 0) ? x * temp * temp : (temp * temp) / x)
: temp * temp);
}
template <typename Arg, typename... Args>
void db(Arg&& arg, Args&&... args) {
cout << std::forward<Arg>(arg);
using expander = long long int[];
(void)expander{0, (void(cout << ',' << std::forward<Args>(args)), 0)...};
cout << "\n";
}
void IO_FILE() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
inline void solve() {
long long int n, l, r;
cin >> n >> l >> r;
if (l == n * (n - 1) + 1) {
cout << 1 << " ";
return;
}
long long int st = 1;
long long int cnt = 0;
while (cnt < l) {
cnt += 2 * (n - st);
if (st == n) {
cnt++;
break;
}
st++;
}
st--;
cnt -= 2 * (n - st);
long long int w, f = 1, j = st + 1;
while (cnt < l) {
f ? w = st : w = j++;
f = 1 - f;
cnt++;
}
if (w != st) {
cout << w << " ";
l++;
}
f = 1;
while (j <= n and l <= r) {
f ? cout << st << " " : cout << j++ << " ";
f = 1 - f;
l++;
}
st++;
while (l <= r) {
f = 1;
j = st + 1;
if (st == n) {
break;
}
while (j <= n and l <= r) {
f ? cout << st << " " : cout << j++ << " ";
f = 1 - f;
l++;
}
st++;
}
if (r == n * (n - 1) + 1) {
cout << 1 << " ";
}
cout << "\n";
}
int32_t main() {
IO_FILE();
long long int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
// package EducationalRound85;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class ProblemD {
public static void main(String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int test=Integer.parseInt(br.readLine());
StringBuilder print=new StringBuilder();
while(test--!=0){
StringTokenizer st=new StringTokenizer(br.readLine());
int n=Integer.parseInt(st.nextToken());
long l=Long.parseLong(st.nextToken());
long r=Long.parseLong(st.nextToken());
long sum[]=new long[n+1];
build(sum);
int ind=search(sum,l,n);
if(ind==-1){
print.append("1\n");
continue;
}
long rem=l-sum[ind-1];
long end=(1l*n*(n-1))+1;
// System.out.println(end);
if(rem%2==0){
long p=ind;
long curr=p+rem/2;
boolean chance=false;
while(l<=r){
if(l==end){
print.append("1");
break;
}
if(chance){
print.append(p+" ");
chance=false;
}
else{
print.append(curr+" ");
chance=true;
curr++;
if(curr>n){
p++;
curr=p+1;
}
}
l++;
}
}
else{
long p=ind;
long curr=p+rem/2+1;
boolean chance=true;
while(l<=r){
if(l==end){
print.append("1");
break;
}
if(chance){
print.append(p+" ");
chance=false;
}
else{
print.append(curr+" ");
chance=true;
curr++;
if(curr>n){
p++;
curr=p+1;
}
}
l++;
}
}
print.append("\n");
}
System.out.print(print.toString());
}
public static void build(long sum[]){
int n=sum.length-1;
for(int i=1;i<=n;i++){
long curr=2l*(n-i);
sum[i]=sum[i-1]+curr;
}
return;
}
public static int search(long sum[],long l,int n){
int ans=-1;
int low=1,high=n-1;
while(low<=high){
int mid=(low+high)/2;
if(sum[mid]>=l){
ans=mid;
high=mid-1;
}
else{
low=mid+1;
}
}
return ans;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
# -*- coding: utf-8 -*-
import sys
from itertools import accumulate
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10 ** 9 + 7
def gen_arr(v):
if N - v == 0:
return [1]
n = (N-v) * 2
res = [0] * n
x = v + 1
for i in range(n):
if i % 2 == 0:
res[i] = v
else:
res[i] = x
x += 1
return res
for _ in range(INT()):
N, l, r = MAP()
l -= 1
ans = []
cur = 0
v = 1
incr = (N-1) * 2
while cur + incr < l:
cur += incr
v += 1
incr -= 2
ans = gen_arr(v)
ans = ans[l-cur:]
ln = r - l
v += 1
while len(ans) < ln:
ans += gen_arr(v)
v += 1
ans = ans[:ln]
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const double pi = acos(-1.0);
const long long int inf = 0x3f3f3f3f3f3f3f3f;
const long long int mod = 998244353;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
void fast() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
long long int power(long long int x, long long int y, long long int p) {
long long int res = 1;
x = x % p;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long int modInverse(long long int n, long long int p) {
return power(n, p - 2, p);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) {
long long int n, l, r;
cin >> n >> l >> r;
long long int k1 = 0, k2 = 0, i1, i2;
for (i2 = 1; i2 < n; i2++) {
k2 += 2 * (n - i2);
if (k2 >= r) break;
}
for (i1 = 1; i1 < n; i1++) {
k1 += 2 * (n - i1);
if (k1 >= l) break;
}
i2++;
vector<long long int> v, a;
k1 -= (n - i1) * 2;
for (int i = i1; i < i2; i++) {
for (int j = i + 1; j < n + 1; j++) {
a.emplace_back(i);
a.emplace_back(j);
}
}
for (long long int i = l; i <= r; i++) {
if (i == (n * (n - 1) + 1))
cout << "1 ";
else
cout << a[i - k1 - 1] << " ";
}
cout << '\n';
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
// package com.company;
import java.util.*;
import java.lang.*;
import java.io.*;
//****Use Integer Wrapper Class for Arrays.sort()****
public class CH4 {
public static void main(String[] Args){
FastReader scan=new FastReader();
int t=scan.nextInt();
StringBuilder fp=new StringBuilder();
while(t-->0){
long n=scan.nextInt();
long l=scan.nextLong();
long r=scan.nextLong();
StringBuilder print=new StringBuilder();
long cur=0;
boolean lf=false;
long ib=-1;
for(long i=1;i<n;i++){
long tba=(n-i)*2;
if(cur+tba<l){
cur+=tba;
}else{
// System.out.println(i);
long fa=l-cur;
if(fa%2!=0&&l<=r){
print.append(i+" ");
fa++;
l++;
}
if(fa%2==0&&l<=r){
print.append((i+fa/2)+" ");
l++;
}
long rem=(tba-fa)/2;
for(long j=n-rem+1;j<=n&&l<=r;j++){
print.append(i+" ");
l++;
if(l<=r){
print.append(j+" ");
l++;
}
}
ib=i+1;
break;
}
}
if(l<=r){
for(long i=ib;ib!=-1&&i<n&&l<=r;i++){
long j=i+1;
while(j<=n){
if(l<=r){
print.append(i+" ");
l++;
}
if(l<=r){
print.append(j+" ");
l++;
}
j++;
}
}
if(l<=r){
print.append(1);
}
}
fp.append(print+"\n");
}
System.out.println(fp);
}
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;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
public static void process(int test_number)throws IOException
{
long n = nl(), l = nl(), r = nl(), termNum = 1;
for(long start = 1l; start <= n - 1 && termNum <= r; start++){
if(termNum + (n - start) * 2l - 1l < l){
termNum += (n - start) * 2l;
continue;
}
for(long j = start + 1l; j <= n && termNum <= r; j++){
if(termNum >= l && termNum <= r)
p(start + " ");
++termNum;
if(termNum >= l && termNum <= r)
p(j + " ");
++termNum;
}
}
if(termNum >= l && termNum <= r)
p("1 ");
p('\n');
}
static final long mod = (long)1e9+7l;
static boolean DEBUG = true;
static FastReader sc;
static PrintWriter out;
public static void main(String[]args)throws IOException
{
out = new PrintWriter(System.out);
sc = new FastReader();
long s = System.currentTimeMillis();
int t = 1;
t = ni();
for(int i = 1; i <= t; i++)
process(i);
out.flush();
System.err.println(System.currentTimeMillis()-s+"ms");
}
static void trace(Object... o){ if(!DEBUG) return; System.err.println(Arrays.deepToString(o)); };
static void pn(Object o){ out.println(o); }
static void p(Object o){ out.print(o); }
static int ni()throws IOException{ return Integer.parseInt(sc.next()); }
static long nl()throws IOException{ return Long.parseLong(sc.next()); }
static double nd()throws IOException{ return Double.parseDouble(sc.next()); }
static String nln()throws IOException{ return sc.nextLine(); }
static long gcd(long a, long b){ return (b==0)?a:gcd(b,a%b);}
static int gcd(int a, int b){ return (b==0)?a:gcd(b,a%b); }
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();
}
String nextLine(){
String str = "";
try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); }
return str;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
int I_INF = 2e9;
long long int L_INF = 1e18;
void solve() {
long long int n, i, j, l, r, index, ctr, num;
cin >> n >> l >> r;
vector<long long int> vect(n, 0);
vect[0] = 2 * (n - 1);
for (i = 1; i < n; i++) {
vect[i] = 2 * (n - 1 - i);
vect[i] += vect[i - 1];
}
vect[n - 1]++;
index =
(long long int)(lower_bound(vect.begin(), vect.end(), l) - vect.begin());
ctr = (index > 0) ? vect[index - 1] : 0;
while (ctr < r) {
if (index == (n - 1)) {
ctr++;
if (ctr >= l && ctr <= r) cout << 1 << " ";
} else {
for (i = index + 1; i < n; i++) {
ctr++;
num = index + 1;
if (ctr >= l && ctr <= r) cout << num << " ";
ctr++;
num = i + 1;
if (ctr >= l && ctr <= r) cout << num << " ";
}
}
index++;
}
cout << endl;
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import io
import os
from collections import Counter, defaultdict, deque
def solve(N, L, R):
M = R - L + 1
ans = []
count = 0
done = False
for i in range(N - 1, -1, -1):
if count + 2 * i < L:
count += 2 * i
else:
curr = N - i
for j in range(curr + 1, N + 1):
if count + 1 >= L:
ans.append(str(curr))
if len(ans) == M:
done = True
break
count += 1
if count + 1 >= L:
ans.append(str(j))
if len(ans) == M:
done = True
break
count += 1
if done:
break
if len(ans) < M:
ans.append("1")
return " ".join(ans)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
T = int(input())
for t in range(T):
N, L, R = [int(x) for x in input().split()]
ans = solve(N, L, R)
print(ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
//package er85;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
public class fourth {
static int n;
// static StringBuilder f(long l, long, r, StringBuilder sb) {
// }
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
int T = in.nextInt();
for (int i=0;i<T;i++) {
n = in.nextInt();
long l = in.nextLong()-1;
long r = in.nextLong();
StringBuilder ans = new StringBuilder();
long index = 0L;
for (int j=n-1;j>=1;j--) {
long a = index;
long b = index+2*j;
if (b<=l || r<=a) {
} else {
int[] nums = new int[2*j];
for (int k=0;k<j;k++) {
nums[2*k] = n-j;
nums[2*k+1] = n-j+k+1;
}
for (long k=Math.max(l, a);k<Math.min(r, b);k++) {
ans.append(nums[(int)(k-a)]);
ans.append(" ");
}
}
index += 2*j;
}
if (r==(long)n*(n-1)+1) {
ans.append(1);
ans.append(" ");
}
ans.deleteCharAt(ans.length()-1);
out.println(ans);
// StringBuilde sb = new StringBuilder();
// out.println(f(l, r, sb));
}
out.close();
}
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());
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
for _ in range(val()):
n,l,r = li()
orig = r
l -= 1
r -= l
currleft = 0
curs = n - 1
while curs and currleft + 2*curs <= l:
currleft += 2*curs
curs -= 1
start = n - curs
ans = []
head = start
l -= currleft
last = head + 1
half = 0
while l:
half = 1
start = last
last += 1
l -= 1
if not l:break
half = 0
start = head
l -= 1
if half:
ans.append(start)
if last == n + 1:
head += 1
last = head + 1
start = head
r -= 1
while r:
ans.append(start)
r -= 1
if not r:break
start = last
ans.append(start)
last += 1
if last == n + 1:
head += 1
last = head + 1
r -= 1
start = head
if orig == n*(n-1) + 1:
ans[-1] = 1
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException
{
FastScanner f = new FastScanner();
int t=1;
t=f.nextInt();
PrintWriter out=new PrintWriter(System.out);
for(int tt=0;tt<t;tt++) {
int n=f.nextInt();
long L=f.nextLong()-1;
long R=f.nextLong()-1;
long base = 0;
for(int i = 1;i <= n-1;i++){
long len = 2*(n-i)-1;
if(Math.max(base, L) <= Math.min(base+len-1, R)){
for(int j = 0;j < len;j++){
if(L <= base+j && base+j <= R){
if(j % 2 == 0){
out.print(i + " ");
}else{
out.print(j/2+i+1 + " ");
}
}
}
}
base += len;
if(L <= base && base <= R){
out.print(n + " ");
}
base++;
}
if(L <= base && base <= R){
out.print(1 + " ");
}
base++;
out.println();
}
out.close();
}
static void sort(int [] a) {
ArrayList<Integer> q = new ArrayList<>();
for (int i: a) q.add(i);
Collections.sort(q);
for (int i = 0; i < a.length; i++) a[i] = q.get(i);
}
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());
}
double nextDouble() {
return Double.parseDouble(next());
}
long[] readLongArray(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
}
//Some things to notice
//Check for the overflow
//Binary Search
//Bitmask
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long n, long long m) {
return n % m == 0 ? m : gcd(m, n % m);
}
long long getNum(long long x, long long n, long long pre, long long k) {
if (k >= n) return 1;
if ((x - pre) & 1)
return k;
else
return k + (x - pre) / 2;
}
int main() {
ios::sync_with_stdio(false);
int T;
cin >> T;
while (T--) {
long long n, l, r;
cin >> n >> l >> r;
long long pre = 0;
long long Next = 2 * (n - 1);
long long k = 1;
for (long long i = l; i <= r; i++) {
while (i > Next && k != n) {
pre = Next;
Next += 2 * (n - k - 1);
k++;
}
cout << getNum(i, n, pre, k) << " ";
}
cout << endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
# HEY STALKER
for _ in range(int(input())):
n, l, r = map(int, input().split())
z = 0
idx = 0
sd = (n-1)*2
lst = (n*(n-1))+1
while z < l:
if sd <= 0:
z += 1
break
idx += 1
z += sd
sd -= 2
m = idx-1
c = (n-1)*2
sm = 0
for t in range(m):
sm += c
c -= 2
sm += 1
nikal = l-sm
p = []
if idx == 1:
for t in range(2, n+1):
p.append(1)
p.append(t)
else:
ii = idx
for t in range(ii+1, n+1):
p.append(ii)
p.append(t)
p.reverse()
for tg in range(nikal):
p.pop()
p.reverse()
idx += 1
i1 = idx
while len(p) < (r-l+1):
if i1 == n:
p.append(1)
for t in range(i1+1, n+1):
p.append(i1)
p.append(t)
i1 += 1
print(*p[:(r-l+1)])
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using namespace std;
long long int gcd(long long int a, long long int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
long long int *y) {
if (a == 0) {
*x = 0;
*y = 1;
return b;
}
long long int x1, y1;
long long int gcd = gcdExtended(b % a, a, &x1, &y1);
*x = y1 - (b / a) * x1;
*y = x1;
return gcd;
}
long long int mod = 1000000007;
long long int binpower(long long int a, long long int b) {
long long int ans = 1;
while (b > 0) {
if (b & 1) ans = (ans * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return ans;
}
bool isPrime(int n) {
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) return false;
}
return true;
}
long long int fact[1005];
void factcalc() {
fact[0] = 1;
fact[1] = 1;
for (long long int i = 2; i < 1005; ++i) {
fact[i] = (fact[i - 1] * i) % mod;
}
}
long long int C(int u, int v) {
long long int c = (fact[v] * fact[u - v]) % mod;
long long int in = binpower(c, mod - 2);
long long int res = (in * fact[u]) % mod;
return res;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long int test = 1;
cin >> test;
while (test--) {
int n;
long long int l, r;
cin >> n >> l >> r;
long long int curr = 0;
bool fl = false;
for (int i = 0; i < n - 1; ++i) {
long long int adder = 0;
if (i == 0)
adder = 2 * (n - 1);
else
adder = 2 * (n - i - 1);
if (curr + adder >= l) {
fl = true;
long long int surplus = (l - curr);
long long int beg = surplus / 2;
int f = i + 1, s = beg + i + 2;
if (surplus % 2 == 0) {
cout << i + 1 + beg << " ";
if (surplus == adder) {
f++;
s = f + 1;
}
l++;
}
int turn = 0;
while (l <= r) {
if (turn == 0) {
if (f == n) {
cout << 1;
break;
}
cout << f << " ";
turn = 1;
l++;
continue;
} else {
cout << s << " ";
s++;
turn = 0;
if (s == n + 1) {
f++;
s = f + 1;
}
l++;
continue;
}
}
break;
}
curr += adder;
}
if (!fl) cout << 1;
cout << endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class MinimumEulerCycle implements Closeable {
private InputReader in = new InputReader(System.in);
private PrintWriter out = new PrintWriter(System.out);
public void solve() {
int T = in.ni();
while (T-- > 0) {
long n = in.nl(), l = in.nl() - 1, r = in.nl() - 1;
long row = findRow(n, l);
long rem = n - row;
long firstIndex = n * (n - 1) - rem * (rem - 1);
long dist = (l - firstIndex + (l % 2)) / 2;
long even = row + 1;
long odd = row + dist + 2 - (l % 2);
if (even == n) {
even = 1;
}
boolean printRow = l % 2 == 0;
for (int i = 0; i < r - l + 1; i++) {
if (printRow) {
out.print(even);
} else {
out.print(odd);
if (odd == n) {
even++;
if (even == n) {
even = 1;
}
odd = even + 1;
} else {
odd++;
}
}
out.print(' ');
printRow = !printRow;
}
out.println();
}
}
private long findRow(long n, long idx) {
long left = 1, right = n - 1, result = 0;
while (left <= right) {
long mid = left + (right - left) / 2;
long rem = n - mid;
long nodes = n * (n - 1) - rem * (rem - 1);
if (nodes <= idx) {
left = mid + 1;
result = Math.max(result, mid);
} else {
right = mid - 1;
}
}
return result;
}
@Override
public void close() throws IOException {
in.close();
out.close();
}
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 ni() {
return Integer.parseInt(next());
}
public long nl() {
return Long.parseLong(next());
}
public void close() throws IOException {
reader.close();
}
}
public static void main(String[] args) throws IOException {
try (MinimumEulerCycle instance = new MinimumEulerCycle()) {
instance.solve();
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long n;
cin >> n;
long long l, r;
cin >> l >> r;
if (l == r && l == (n * (n - 1) + 1)) {
cout << 1 << '\n';
return;
}
long long grp = n - 1;
long long sum = 0;
long long idx = 1;
long long extra = 0;
for (long long i = 1; i <= n - 1; i++) {
sum += (grp * 2);
if (sum >= l) {
idx = i;
sum -= (grp * 2);
break;
}
grp--;
}
long long rem = l - sum;
long long x = idx;
long long y = idx + (rem + 1) / 2;
for (long long i = l; i <= r; i++) {
if (i == (n * (n - 1) + 1)) {
cout << 1 << ' ';
} else {
if (i % 2) {
cout << x << ' ';
} else {
cout << y << ' ';
y++;
if (y == n + 1) {
x++;
y = 1 + x;
}
}
}
}
cout << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
int t = 1;
cin >> t;
for (int i = 1; i <= t; i++) {
solve();
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int ar[200002];
int main() {
ios::sync_with_stdio(0);
cin.tie(nullptr);
cout.tie(nullptr);
;
long long t, n, l, r;
cin >> t;
while (t--) {
cin >> n >> l >> r;
if (l == n * (n - 1) + 1) {
cout << "1\n";
continue;
}
long long s = 0, j = 1;
while (s + 2 * (n - j) < l) {
s += 2 * (n - j);
j++;
}
long long skipped = (l - s - 1) / 2;
long long start = j, x = j + skipped + 1;
;
long long end = n * (n - 1) + 1;
for (long long a = l; a <= min(r, end - 1); a++) {
if (a % 2) {
cout << start << " ";
} else {
cout << x++ << " ";
}
if (x > n) {
start++;
x = start + 1;
}
}
if (r == end) {
cout << 1;
}
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.StringTokenizer;
import java.util.function.*;
import java.util.function.Function;
import java.util.function.Predicate;
import java.util.stream.*;
public class D {
private static final FastReader in = new FastReader();
private static final FastWriter out = new FastWriter();
public static void main(String[] args) {
new D().run();
}
private void run() {
var t = in.nextInt();
while (t-- > 0) {
solve();
}
out.flush();
}
int n;
long[] sum;
private void solve() {
n = in.nextInt();
var l = in.nextLong();
var r = in.nextLong();
sum = new long[n];
for (var i = 1; i < n; i++) {
sum[i] = sum[i - 1] + (n - i) * 2;
}
var ans = new long[(int) (r - l + 1)];
for (var i = l; i <= r; i++) {
ans[(int) (i - l)] = euler(i);
}
out.println(ans);
}
long euler(long i) {
if (i > sum[n - 1]) return 1;
var x = Misc.lowerBound(sum, i);
var s = sum[x - 1];
var d = i - s;
return d % 2 == 1 ? x : x + d / 2;
}
}
class FastReader {
private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private static StringTokenizer in;
public String next() {
while (in == null || !in.hasMoreTokens()) {
try {
in = new StringTokenizer(br.readLine());
} catch (IOException e) {
return null;
}
}
return in.nextToken();
}
public BigDecimal nextBigDecimal() {
return new BigDecimal(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
public boolean nextBoolean() {
return Boolean.valueOf(next());
}
public byte nextByte() {
return Byte.valueOf(next());
}
public double nextDouble() {
return Double.valueOf(next());
}
public double[] nextDoubleArray(int length) {
var a = new double[length];
for (var i = 0; i < length; i++) {
a[i] = nextDouble();
}
return a;
}
public int nextInt() {
return Integer.valueOf(next());
}
public int[] nextIntArray(int length) {
var a = new int[length];
for (var i = 0; i < length; i++) {
a[i] = nextInt();
}
return a;
}
public long nextLong() {
return Long.valueOf(next());
}
public long[] nextLongArray(int length) {
var a = new long[length];
for (var i = 0; i < length; i++) {
a[i] = nextLong();
}
return a;
}
}
class FastWriter extends PrintWriter {
public FastWriter() {
super(System.out);
}
public void println(double[] a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public void println(int[] a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public void println(long[] a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public void println(Object... a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public <T> void println(List<T> l) {
println(l.toArray());
}
public void debug(String name, Object o) {
String value = Arrays.deepToString(new Object[] { o });
value = value.substring(1, value.length() - 1);
System.err.println(name + " => " + value);
}
}
class Misc {
public static final double EPS = 1e-12;
public static final Comparator<Double> EPS_COMPARATOR = (x, y) -> {
if (x + EPS < y) {
return -1;
} else if (x - EPS > y) {
return 1;
} else {
return 0;
}
};
public static int compare(double x, double y) {
return EPS_COMPARATOR.compare(x, y);
}
/**
* Returns the index of the first element in the range <b>[left, right)</b> which <i>leftShouldAdvance</i> tested to
* be <i>false</i>.
*/
public static int binarySearch(int left, int right, Predicate<Integer> leftShouldAdvance) {
while (left < right) {
var mid = left + (right - left) / 2;
if (leftShouldAdvance.test(mid)) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
/**
* Returns the index of the first element in <i>a</i> which >= <i>x</i>.
*/
public static int lowerBound(int[] a, int x) {
return binarySearch(0, a.length, mid -> a[mid] < x);
}
public static int lowerBound(long[] a, long x) {
return binarySearch(0, a.length, mid -> a[mid] < x);
}
/**
* Returns the index of the first element in <i>a</i> which > <i>x</i>.
*/
public static int upperBound(int[] a, int x) {
return binarySearch(0, a.length, mid -> a[mid] <= x);
}
public static int upperBound(long[] a, long x) {
return binarySearch(0, a.length, mid -> a[mid] <= x);
}
/**
* Searches for the maximum value of a unimodal function f(x).
* <p>
* A function f(x) is a <b>unimodal function</b> if for some value m, it is <b>monotonically increasing</b> for x β€
* m and <b>monotonically decreasing</b> for x β₯ m. In that case, the maximum value of f(x) is f(m) and there are no
* other local maxima.
*/
public static int ternarySearch(int left, int right, Function<Integer, Integer> f) {
return binarySearch(left, right, mid -> f.apply(mid) < f.apply(Math.min(mid + 1, right - 1)));
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int t;
cin >> t;
while (t--) {
solve();
}
}
const long long N = 1e5 + 5;
std::vector<long long> cnt(N, 0);
void solve() {
long long n, l, r;
cin >> n >> l >> r;
for (long long i = 1; i < n + 1; ++i) {
cnt[i] = 2 * (n - i);
}
for (long long i = 2; i < n + 1; ++i) {
cnt[i] += cnt[i - 1];
}
long long p = 0;
for (long long i = l; i <= r; ++i) {
while (p + 1 <= n and cnt[p + 1] < i) {
p++;
}
long long temp = i - cnt[p];
if (i == ((n * (n - 1)) + 1)) {
cout << 1 << " ";
} else if (i % 2) {
cout << p + 1 << " ";
continue;
} else {
cout << p + 1 + (temp / 2) << " ";
}
}
cout << '\n';
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
t = int(input())
for case_num in range(t):
n, l, r = map(int, input().split(' '))
if l == n * (n-1) + 1:
print(1)
continue
total = 0
unvisited = n - 1
while total < l:
total += unvisited * 2
unvisited -= 1
unvisited += 1
total -= unvisited * 2
current = n - unvisited
nxt = current + 1
ans = []
while total < r:
total += 1
now = current if total % 2 == 1 else nxt
if now == nxt:
nxt += 1
if nxt > n:
current += 1
nxt = current + 1
if current == n:
current = 1
if total >= l:
ans.append(now)
print(' '.join(map(str, ans)))
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
import java.util.stream.Collectors;
public class Main {
long f(int n, int h){
return (long) 2 * h * n - (long) h * (h + 1);
}
void solve2() {
int n = in.nextInt();
long l = in.nextLong();
long r = in.nextLong();
int h = 1;
while(f(n,h)<l && h<n-1) h++;
long tmp = l-f(n,h-1);
boolean tt=tmp%2==1;
long s = h+(tmp+1)/2;
if (s==n+1){
h = n;
}
r++;
int fh = 1;
while(f(n,fh)<r && fh<n-1) fh++;
tmp = r-f(n,fh-1);
boolean ftt=tmp%2==1;
long fs = fh+(tmp+1)/2;
if (fs==n+1){
fh = n;
}
while(true){
if (h==fh && tt==ftt && s==fs){
break;
}
if (h == n) {
out.print(1);
break;
}
if (tt) {
out.print(h+" ");
tt = false;
}else {
out.print(s+" ");
s++;
if (s == n + 1) {
h++;
s = h + 1;
}
tt = true;
}
}
out.println();
}
void solve() {
int t = in.nextInt();
for(int i=0;i<t;i++){
solve2();
}
}
// --------------------SCANNER-------------------------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner(boolean debug) {
if (debug) {
try {
br = new BufferedReader(new FileReader("input.txt"));
} catch (FileNotFoundException e) {
throw new RuntimeException(e);
}
} else {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(nextLine());
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextInts(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
try {
String line = br.readLine();
if (line == null) {
throw new RuntimeException("empty line");
}
return line;
} catch (IOException e) {
throw new RuntimeException(e);
}
}
}
// --------------------WRITER-------------------------
public static class MyWriter extends PrintWriter {
public MyWriter(OutputStream out) {
super(out);
}
void println(int[] arr) {
String line = Arrays.stream(arr).mapToObj(String::valueOf).collect(Collectors.joining(" "));
println(line);
}
}
// --------------------MAIN-------------------------
public MyScanner in;
public MyWriter out;
public static void main(String[] args) {
Main m = new Main();
m.in = new MyScanner(args.length > 0);
m.out = new MyWriter(new BufferedOutputStream(System.out));
m.solve();
m.out.close();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import sys
input = sys.stdin.readline
T=int(input())
for tests in range(T):
n,l,r=map(int,input().split())
begin=1
while l>(n-begin)*2+1:
if begin==n:
break
l-=(n-begin)*2
r-=(n-begin)*2
begin+=1
#print(begin,l,r)
if begin==n:
ANS=[n,1]
else:
ANS=[]
while len(ANS)<=r:
if begin==n:
ANS.append(1)
break
for j in range(begin+1,n+1):
ANS.append(begin)
ANS.append(j)
begin+=1
#print(ANS)
print(*ANS[l-1:r])
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Task {
public static void main(String[] args) throws Exception {
new Task().go();
}
PrintWriter out;
Reader in;
BufferedReader br;
Task() throws IOException {
try {
//br = new BufferedReader( new FileReader("input.txt") );
//in = new Reader("input.txt");
in = new Reader("input.txt");
out = new PrintWriter( new BufferedWriter(new FileWriter("output.txt")) );
}
catch (Exception e) {
//br = new BufferedReader( new InputStreamReader( System.in ) );
in = new Reader();
out = new PrintWriter( new BufferedWriter(new OutputStreamWriter(System.out)) );
}
}
void go() throws Exception {
//int t = in.nextInt();
int t = 1;
while (t > 0) {
solve();
t--;
}
out.flush();
out.close();
}
int inf = 2000000000;
int mod = 1000000007;
double eps = 0.000000001;
int n;
int m;
ArrayList<Integer>[] g;
void solve() throws IOException {
int t = in.nextInt();
while (t > 0) {
int n = in.nextInt();
long l = in.nextLong();
long r = in.nextLong();
long cur = 1;
int x = n - 1;
int first = 1;
while (x > 0 && cur + x * 2 <= l) {
cur += x * 2;
first++;
x--;
}
if (first == n) {
first = 1;
}
int flag = 0;
int lv = first;
int cnt = first + 1;
while (cur <= r) {
if (cur >= l)
out.print(lv + " ");
if (flag == 1) {
flag = 0;
cnt++;
if (cnt > n) {
first++;
cnt = first + 1;
if (first == n) first = 1;
}
lv = first;
} else {
lv = cnt;
flag = 1;
}
cur++;
}
out.println();
t--;
}
}
class Pair implements Comparable<Pair> {
int a;
int b;
Pair(int a, int b) {
this.a = a;
this.b = b;
}
public int compareTo(Pair p) {
if (a != p.a)
return Integer.compare(a, p.a);
else
return Integer.compare(-b, -p.b);
}
}
class Item {
int a;
int b;
int c;
Item(int a, int b, int c) {
this.a = a;
this.b = b;
this.c = c;
}
}
class Reader {
BufferedReader br;
StringTokenizer tok;
Reader(String file) throws IOException {
br = new BufferedReader( new FileReader(file) );
}
Reader() throws IOException {
br = new BufferedReader( new InputStreamReader(System.in) );
}
String next() throws IOException {
while (tok == null || !tok.hasMoreElements())
tok = new StringTokenizer(br.readLine());
return tok.nextToken();
}
int nextInt() throws NumberFormatException, IOException {
return Integer.valueOf(next());
}
long nextLong() throws NumberFormatException, IOException {
return Long.valueOf(next());
}
double nextDouble() throws NumberFormatException, IOException {
return Double.valueOf(next());
}
String nextLine() throws IOException {
return br.readLine();
}
}
static class InputReader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public InputReader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public InputReader(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')
break;
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();
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
template <typename T, typename U>
static inline void amin(T& x, U y) {
if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T& x, U y) {
if (x < y) x = y;
}
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long solve() {
long long n, l, r;
cin >> n >> l >> r;
if (n == 2) {
long long a[] = {1, 2, 1};
for (long long i = l - 1; i < r; i++) cout << a[i] << " ";
cout << '\n';
return 0;
}
long long z = 0;
for (long long i = 1; i < n; i++) {
long long cnt = 2 * (n - i);
if (z + cnt + 1 < l) {
z += cnt;
continue;
}
if (l > r) break;
if (l == z + 1 && l <= r) {
cout << i << " ";
l++;
}
z++;
for (long long j = i + 1; j < n; j++) {
if (l == z + 1 && l <= r) {
cout << j << " ";
l++;
}
z++;
if (l == z + 1 && l <= r) {
cout << i << " ";
l++;
}
z++;
}
if (l == z + 1 && l <= r) {
cout << n << " ";
l++;
}
z++;
}
if (l <= r && l == z + 1) cout << 1 << " ";
return 0;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
x = 0;
bool f = 0;
char c = getchar();
for (; !isdigit(c); c = getchar())
if (c == '-') f = 1;
for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
if (f) x = -x;
}
template <typename T>
inline void Mx(T &x, T y) {
x < y && (x = y);
}
template <typename T>
inline void Mn(T &x, T y) {
x > y && (x = y);
}
long long T, n, l, r;
int main() {
read(T);
while (T--) {
read(n), read(l), read(r);
if (n == 2) {
}
long long sum = 0, pre = 0;
for (int i = 1; i < n; i++) {
sum += (n - i) * 2;
if (l < sum) {
long long t = l - pre;
long long x = (t & 1) ? i : (t / 2) + i, y = (t / 2) + i;
while (l < sum && l <= r) {
printf("%lld ", x);
x == i ? y++, x = y : x = i;
l++;
}
}
if (l == sum && l <= r) printf("%lld ", n), l++;
pre = sum;
if (l > r) break;
}
if (l <= r) printf("1");
puts("");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
# coding: utf-8
# Your code here!
def solve():
n, l, r = map(int, input().split())
def where(x):
if x == n*(n-1) + 1:
return 1
else:
l = 0
r = n + 1
while r - l > 1:
m = (l + r)//2
if (2*n-1-m)*(m) < x:
l = m
else:
r = m
v = x - (2*n-1-l)*(l)
if v % 2 != 0:
return r
else:
return r + v//2
res = [where(p) for p in range(l, r+1)]
print(*res)
return
def main():
t = int(input())
for i in range(t):
solve()
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
# region fastio # from https://codeforces.com/contest/1333/submission/75948789
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
T = int(input())
for idx_testcase in range(T):
N, L, R = map(int, input().split())
f = False
if R==N*(N-1)+1:
if L==R:
print(1)
continue
f = True
R -= 1
s = 0
d = 2*N
i = 0
Ans = [0] * (R-L+1)
while s < L:
d -= 2
s += d
i += 1
s -= d
numer = L-s
for idx_ans in range(R-L+1):
if numer%2:
Ans[idx_ans] = i
else:
Ans[idx_ans] = numer//2 + i
numer += 1
if Ans[idx_ans] == N:
numer = 1
i += 1
if f:
Ans.append(1)
print(*Ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import sys
input = sys.stdin.readline
from collections import *
def binary_search1():
l, r = 1, n-1
while l<=r:
m = (l+r)//2
if acc[m]-acc[1]+1<=left:
l = m+1
else:
r = m-1
return r
def binary_search2():
l, r = 1, n-1
while l<=r:
m = (l+r)//2
if acc[m+1]-acc[1]>=right:
r = m-1
else:
l = m+1
return l
T = int(input())
for _ in range(T):
n, left, right = map(int, input().split())
l = [0, 2*n-2]
for _ in range(n-2):
l.append(l[-1]-2)
l[-1] += 1
#print(l)
acc = [0]
for li in l:
acc.append(acc[-1]+li)
left_n = binary_search1()
right_n = binary_search2()
#print(left_n)
#print(right_n)
ans = []
for i in range(left_n, right_n+1):
li = []
for j in range(l[i]-1):
if j%2==0:
li.append(i)
else:
if j==1:
li.append(i+1)
else:
li.append(li[-2]+1)
if i==n-1:
li.append(1)
else:
li.append(n)
ans += li
#print(ans)
sta = left-(acc[left_n]-acc[1])-1
#print(sta)
print(*ans[sta:sta+right-left+1])
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
TaskD solver = new TaskD();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class TaskD {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int N = in.nextInt();
long L = in.nextLong(), R = in.nextLong();
long index = 1, increment = 2 * (N - 1), num = 1, right = -1;
while (index < L) {
if (index + increment <= L) {
index += increment;
increment -= 2;
num++;
} else {
right = num + 1;
while (index < L) {
if (L - index >= 2) {
index += 2;
right++;
} else {
index += 2;
out.print(right + " ");
right++;
}
}
for (long i = right; i <= N; i++) {
if (index > R) {
out.printLine();
return;
}
out.print(num + " ");
index++;
if (index > R) {
out.printLine();
return;
}
out.print(i + " ");
index++;
}
num++;
break;
}
}
for (long i = num; i < N; i++) {
for (long j = i + 1; j <= N; j++) {
if (index > R) {
out.printLine();
return;
}
out.print(i + " ");
index++;
if (index > R) {
out.printLine();
return;
}
out.print(j + " ");
index++;
}
}
if (index <= R) {
out.print("1");
}
out.printLine();
}
}
static class OutputWriter {
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();
}
}
static class InputReader {
BufferedReader in;
StringTokenizer tokenizer = null;
public InputReader(InputStream inputStream) {
in = new BufferedReader(new InputStreamReader(inputStream));
}
public String next() {
try {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(in.readLine());
}
return tokenizer.nextToken();
} catch (IOException e) {
return null;
}
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import sys
input = sys.stdin.readline
def li():return [int(i) for i in input().rstrip('\n').split()]
def val():return int(input().rstrip('\n'))
for _ in range(val()):
n,l,r = li()
orig = r
l -= 1
r -= l
currleft = 0
curs = n - 1
while curs and currleft + 2*curs <= l:
currleft += 2*curs
curs -= 1
start = n - curs
ans = []
head = start
l -= currleft
last = head + 1
half = 0
while l:
half = 1
start = last
last += 1
l -= 1
if not l:break
half = 0
start = head
l -= 1
if half:
ans.append(start)
if last == n + 1:
head += 1
last = head + 1
start = head
r -= 1
while r:
ans.append(start)
r -= 1
if not r:break
start = last
ans.append(start)
last += 1
if last == n + 1:
head += 1
last = head + 1
r -= 1
start = head
if orig == n*(n-1) + 1:
ans[-1] = 1
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long int oo = numeric_limits<long long int>::max();
long long int MOD = 1e9 + 7;
long long int comp(long long int n, long long int i) {
return 2 * 1LL * (n - i);
}
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
cin.exceptions(cin.failbit);
int t;
cin >> t;
while (t--) {
long long int n, l, r;
cin >> n >> l >> r;
long long int i = 1;
long long int sm = comp(n, i);
while (i <= n && sm < l) {
i++;
sm += comp(n, i);
}
long long int idx = sm - comp(n, i);
for (; i <= n; i++) {
for (long long int j = 0; j < comp(n, i); j++) {
long long int nm;
if (j % 2 == 0) {
nm = i;
} else {
nm = (j / 2) + i + 1LL;
}
idx++;
if (idx >= l && idx <= r) {
cout << nm << " ";
} else if (idx > r) {
goto end;
}
}
}
idx++;
if (idx <= r) cout << 1;
end:
cout << endl;
continue;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int n;
long long l, r;
long long curlen;
void add(int u) {
curlen++;
if (l <= curlen && curlen <= r) printf("%d ", u);
return;
}
void Add(int u) {
int cnt = (n - u) * 2;
if (curlen + cnt >= l && curlen < r) {
for (int i = u + 1; i <= n; i++) {
add(u);
add(i);
}
} else
curlen += cnt;
return;
}
void solve() {
scanf("%d %I64d %I64d", &n, &l, &r);
curlen = 0;
for (int i = 1; i < n; i++) Add(i);
add(1);
printf("\n");
return;
}
int main() {
int t;
scanf("%d", &t);
while (t--) solve();
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t;
t = 1;
cin >> t;
for (int i1 = 0; i1 < t; ++i1) {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> v;
long long c = (n - 1) * 2;
long long s = c;
vector<long long> v1;
for (int i = 0; i < n - 1; ++i) {
v.push_back(s);
v1.push_back(c / 2);
c -= 2;
s += c;
}
long long x = lower_bound(v.begin(), v.end(), l) - v.begin();
long long y = lower_bound(v.begin(), v.end(), r) - v.begin();
if (x == v.size()) {
cout << 1 << "\n";
continue;
}
long long res1 = v[x] - l + 1;
if (res1 % 2) {
res1 /= 2;
cout << n - res1 << " ";
l++;
} else
res1 /= 2;
if (x == y) {
long long dif = r - l + 1;
while (dif > 1) {
res1--;
cout << x + 1 << " " << n - res1 << " ";
dif -= 2;
}
if (dif == 1) cout << x + 1 << "\n";
continue;
}
for (int i = 0; i < res1; ++i) {
cout << x + 1 << " " << n - res1 + 1 + i << " ";
}
x++;
while (x < y) {
for (int i = 0; i < v1[x]; ++i) {
cout << x + 1 << " " << x + i + 2 << " ";
}
x++;
}
if (y == v.size()) {
cout << 1 << "\n";
continue;
}
long long res2 = r - v[y - 1];
res1 = res2;
res2 /= 2;
for (int i = 0; i < res2; ++i) {
cout << y + 1 << " " << y + 2 + i << " ";
}
if (res1 % 2) cout << y + 1;
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import math
# ΡΠ΅ΡΠ΅Π½Π°
def task_1343_c():
b = int(input())
array = [int(num) for num in input().split()]
maxPositive = 0
minNegative = -10000000000
res = 0
for i in range(b):
if array[i] < 0:
if i != 0 and array[i - 1] >= 0:
res += maxPositive
maxPositive = 0
minNegative = max(minNegative, array[i])
else:
if i != 0 and array[i - 1] < 0:
res += minNegative
minNegative = -10000000000
maxPositive = max(maxPositive, array[i])
if minNegative == -10000000000:
res += maxPositive
else:
res += maxPositive + minNegative
print(res)
# Π½Π΅ ΡΠ°Π±ΠΎΡΠ°Π΅Ρ ΠΎΡ ΡΠ»ΠΎΠ²Π° ΡΠΎΠ²ΡΠ΅ΠΌ
def task_1341_b():
heightLen, doorSize = map(int, input().split())
heights = [int(num) for num in input().split()]
perf = [0 for i in range(heightLen)]
a = 0
for i in range(heightLen - 1):
if i == 0:
perf[i] = 0
else:
if heights[i - 1] < heights[i] and heights[i] > heights[i + 1]:
a += 1
perf[i] = a
perf[heightLen - 1] = a
max_global = 0
left_global = 0
for i in range(heightLen - doorSize):
max_local = perf[i + doorSize - 1] - perf[i]
if max_local > max_global:
max_global = max_local
left_global = i
print(max_global + 1, left_global + 1)
# ΡΠ΅ΡΠΈΠ», ΡΡΠΎΠ± Π΅Ρ
def task_1340_a():
n = int(input())
array = [int(i) for i in input().split()]
for i in range(n - 1):
if array[i] < array[i + 1]:
if array[i] + 1 != array[i + 1]:
print("No")
return
print("Yes")
#ΡΠ΅ΡΠΈΠ»
def task_1339_b():
n = int(input())
array = [int(num) for num in input().split()]
array.sort()
output = [0 for i in range(0, n)]
i = 0
h = 0
j = n - 1
while i <= j:
output[h] = array[i]
h += 1
i += 1
if h < n:
output[h] = array[j]
h += 1
j -= 1
for val in reversed(output):
print(val, end=' ')
# ΡΠ΅ΡΠ΅Π½Π°
def task_1338_a():
n = int(input())
inputArr = [int(num) for num in input().split()]
max_sec = 0
for i in range(1, n):
local_sec = 0
a = inputArr[i - 1] - inputArr[i]
if a <= 0:
continue
else:
b = math.floor(math.log2(a))
local_sec = b + 1
for j in range(b, -1, -1):
if a < pow(2, j):
continue
inputArr[i] += pow(2, j)
a -= pow(2, j)
if local_sec > max_sec:
max_sec = local_sec
print(max_sec)
def task_1334_d():
n, l ,r = map(int, input().split())
if l == 9998900031:
print(1)
return
res = []
count = 0
start_pos = l
for i in range(1, n + 1):
count += (n - i) * 2
if count >= l:
for j in range(n - i):
res.append(i)
res.append(j + i + 1)
else:
start_pos -= (n - i) * 2
if count >= r:
break
res.append(1)
for i in range(start_pos - 1, start_pos + (r - l)):
print(res[i], end=" ")
print()
a = int(input())
for i in range(a):
task_1334_d()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int INF = 2e9;
const long long INFLL = 1e18;
const int MAX_N = 1;
int T;
long long N, L, R;
int main() {
scanf("%d", &T);
while (T--) {
scanf("%lld", &N);
scanf("%lld%lld", &L, &R);
long long n = 1;
while (L <= R) {
while (n != N && L > (N - n) * 2LL) {
L -= (N - n) * 2LL;
R -= (N - n) * 2LL;
n++;
}
if (n == N) {
printf("1");
break;
}
if (L % 2 == 1) {
printf("%lld ", n);
} else {
printf("%lld ", (L / 2) + n);
}
L++;
}
printf("\n");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
def slv(n,l, r):
l-=1
r-=1
c = 0
ans = []
for i in range(1, n):
nl,nr = c, c + 2*(n-i)-1
c = nr+1
if l > nr:
continue
for j in range(max(l, nl), min(r, nr)+1):
if j %2 == 0:
ans.append(i)
else:
ans.append(i + (j-nl+1)//2)
if r == n*(n-1):
ans.append(1)
print(' '.join([str(i) for i in ans]))
t = int(input())
for _ in range(t):
n,l,r = map(int, input().split())
slv(n,l,r)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void display(int lvl, int lt, int rt) {
for (int i = lt; i <= rt; i++)
printf("%d ", ((i & 1) ? lvl : lvl + (i >> 1)));
}
int main() {
int t, n, i, j, k;
cin >> t;
while (t--) {
long long l, r, lvl, sz, sc, mx;
scanf("%d %lld %lld", &n, &l, &r);
mx = n * (n - 1ll) + 1;
if (l == mx) {
printf("1\n");
continue;
}
for (lvl = 1, sz = 0; lvl < n; lvl++) {
sc = 2 * (n - lvl);
if (sz >= l) {
display(lvl, 1, min(sc, r - sz));
} else if (sz + sc >= l) {
display(lvl, l - sz, min(sc, r - sz));
}
sz += sc;
if (sz >= r) {
puts("");
break;
}
}
if (r == mx) {
printf("1\n");
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class C {
public static void main(String[] args) {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-->0) {
long n = in.nextInt(), l = in.nextLong(), r = in.nextLong();
long x = 1;
long sum = 1;
while(sum+(n-x)*2<l){
sum += (n-x)*2;
x++;
}
while(sum<=r){
if(sum>n*(n-1)){
out.print(1); break;
}
for(long i=x+1;i<=n;i++){
if(sum>=l&&sum<=r) out.print(x+" ");
sum++;
if(sum>=l&&sum<=r) out.print(i+" ");
sum++;
}
x++;
}
out.println();
}
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) {}
return st.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
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] = nextInt();
return a;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
from functools import lru_cache #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------
for _ in range(int(input())):
n,l,r = mapi()
low = 1
high = n-1
pst = 0
def check(pst,n,l):
val = 2*n*pst-(pst*pst)-pst
if val < l:
return True
return False
while low <= high:
mid = (low+high)//2
if check(mid,n,l):
pst = mid
low = mid+1
else:
high = mid-1
val = 2*n*pst-(pst*pst)-pst
rem = l-val
#print(pst)
temp = True
prev = True
for i in range(rem-1):
if prev:
if temp:
pst += 1
nxt = pst
else:
nxt += 1
temp = False
if nxt == n:
temp = True
prev = not prev
#print(pst,nxt)
#print(temp,prev,nxt)
res = []
for i in range(r-l+1):
if prev:
if temp:
pst += 1
nxt = pst
if pst == n:
pst = 1
res.append(pst)
else:
temp = False
nxt += 1
res.append(nxt)
if nxt == n:
temp = True
prev = not prev
print(*res)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Main implements Runnable
{
boolean multiple = true;
long MOD;
@SuppressWarnings({"Duplicates", "ConstantConditions"})
void solve() throws Exception
{
long n = sc.nextLong();
long l = sc.nextLong();
long r = sc.nextLong();
long toPrint = r - l + 1;
long curr = 1;
for (long group = 1; group <= n; group++)
{
//how many in this group
long num = 2 * (n - group);
if (group == n && toPrint != 0) { p(1); break; }
if (l >= curr + num) { curr += num; continue; }
// System.out.println(group + " " + toPrint);
//else we are going to start printing
long start = max(curr, l);
// System.out.println(group + " " + start + " " + num + " gay");
for (long idx = start; idx < curr + num; idx++)
{
if (toPrint == 0) break;
long i = idx - curr + 1;
if (i % 2 == 1) p(group);
else p(1 + (i / 2) + (group - 1));
p(' ');
toPrint--;
}
curr += num;
if (toPrint == 0) break;
}
pl();
}
StringBuilder ANS = new StringBuilder();
void p(Object s) { ANS.append(s); } void p(double s) {ANS.append(s); } void p(long s) {ANS.append(s); } void p(char s) {ANS.append(s); }
void pl(Object s) { ANS.append(s); ANS.append('\n'); } void pl(double s) { ANS.append(s); ANS.append('\n'); } void pl(long s) { ANS.append(s); ANS.append('\n'); } void pl(char s) { ANS.append(s); ANS.append('\n'); } void pl() { ANS.append(('\n')); }
/*I/O, and other boilerplate*/ @Override public void run() { try { in = new BufferedReader(new InputStreamReader(System.in));out = new PrintWriter(System.out);sc = new FastScanner(in);if (multiple) { int q = sc.nextInt();for (int i = 0; i < q; i++) solve(); } else solve(); System.out.print(ANS); } catch (Throwable uncaught) { Main.uncaught = uncaught; } finally { out.close(); }} public static void main(String[] args) throws Throwable{ Thread thread = new Thread(null, new Main(), "", (1 << 26));thread.start();thread.join();if (Main.uncaught != null) {throw Main.uncaught;} } static Throwable uncaught; BufferedReader in; FastScanner sc; PrintWriter out; } class FastScanner { BufferedReader in; StringTokenizer st; public FastScanner(BufferedReader in) {this.in = in;}public String nextToken() throws Exception { while (st == null || !st.hasMoreTokens()) { st = new StringTokenizer(in.readLine()); }return st.nextToken(); }public int nextInt() throws Exception { return Integer.parseInt(nextToken()); }public long nextLong() throws Exception { return Long.parseLong(nextToken()); }public double nextDouble() throws Exception { return Double.parseDouble(nextToken()); }
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long ind[100005];
int main() {
ios::sync_with_stdio(0);
cin.tie(NULL);
int t;
cin >> t;
for (int z = 1; z <= t; z++) {
long long n;
long long l, r;
cin >> n >> l >> r;
vector<int> res;
ind[0] = 0;
for (int i = 1; i < n; i++) {
ind[i] = ind[i - 1] + 2 * (n - i);
}
ind[n] = ind[n - 1] + 1;
bool isLast = 0;
long long last = n * (n - 1) + 1;
if (r == last) {
r--;
isLast = 1;
}
int groupL = 0;
for (int i = 1; i <= n; i++) {
if (ind[i] >= l) {
groupL = i;
break;
}
}
int groupR = 0;
for (int i = 1; i <= n; i++) {
if (ind[i] >= r) {
groupR = i;
break;
}
}
long long a = max(ind[groupL - 1] + 1, l);
a -= ind[groupL - 1];
long long b = min(ind[groupL], r);
b -= ind[groupL - 1];
for (long long i = a; i <= b; i++) {
if (i % 2 == 1) {
res.push_back(groupL);
} else {
long long k = i / 2;
k += groupL;
res.push_back(k);
}
}
if (groupR > groupL) {
int group = groupL;
for (long long i = ind[groupL] + 1; i <= ind[groupR - 1]; i++) {
long long k = i - ind[group];
if (k % 2 == 1) {
res.push_back(group + 1);
} else {
long long p = k / 2;
p += group + 1;
res.push_back(p);
if (p == n) {
group++;
}
}
}
long long c = max(ind[groupR - 1] + 1, l);
c -= ind[groupR - 1];
long long d = min(ind[groupR], r);
d -= ind[groupR - 1];
for (long long i = c; i <= d; i++) {
if (i % 2 == 1) {
res.push_back(groupR);
} else {
long long k = i / 2;
k += groupR;
res.push_back(k);
}
}
}
if (isLast == 1) {
res.push_back(1);
}
for (int i = 0; i < res.size(); i++) {
cout << res[i] << " ";
}
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
def genGroup(lo, n):
if lo == n:
return [1]
s = []
for i in range(lo+1, n+1):
s.append(lo)
s.append(i)
return s
for tc in range(int(input())):
n, beg, end = map(int, input().split())
if beg == n*(n-1)+1:
print(1)
else:
past = 0
i = 1
while past + 2*(n-i) < beg:
past += 2*(n-i)
i += 1
#print(i, past)
group = i
s = genGroup(group, n)
pos = beg - past - 1
res = []
for i in range(end-beg+1):
res.append(s[pos])
pos += 1
if pos == len(s):
pos = 0
group += 1
s = genGroup(group, n)
print(*res)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int M = 1e5 + 7;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int T;
cin >> T;
while (T--) {
long long n, l, r, f = 0;
cin >> n >> l >> r;
l--;
if (r == n * (n - 1) + 1) r--, f = 1;
long long nl = (n - 1) * 2, nr = (n - 1) * 2;
while (l >= nl && nl) l -= nl, nl -= 2;
while (r >= nr && nr) r -= nr, nr -= 2;
while (nl >= nr && nl) {
long long tp = n - (nl / 2);
long long nm = 0;
vector<long long> v;
for (long long i = tp + 1; i <= n; i++) v.push_back(tp), v.push_back(i);
if (nl == nr)
for (int i = l; i < r; i++) cout << v[i] << " ";
else {
for (int i = l; i < v.size(); i++) cout << v[i] << " ";
l = 0;
}
nl -= 2;
}
if (f) cout << 1;
cout << endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
#code
import sys
import math as mt
#input=sys.stdin.buffer.readline
t=int(input())
#tot=0
for __ in range(t):
#n=int(input())
#l=list(map(int,input().split()))
n,l,r=map(int,input().split())
j=1
k=2*n-2
mul=1
k=2*n-2
r1=k
l1=1
for i in range(n-2):
if l>=l1 and l<=r1:
#print(111,l1,r1,mul)
break
k-=2
l1=r1+1
r1=l1+k-1
mul+=1
#print(111,l1,r1,mul,k)
nex=mul
ch=0
#print(999,ch)
for i in range(l1,min(r1+1,r+1)):
if i>=l:
if ch%2!=0:
nex+=1
print(nex,end=" ")
else:
print(mul,end=" ")
else:
if ch%2!=0:
nex+=1
ch+=1
if i>r1:
break
i=r1+1
mul+=1
ch=0
nex=mul
k-=1
while i<=min(r,n*(n-1)):
if ch%2==0:
print(mul,end=" ")
else:
nex+=1
print(nex,end=" ")
i+=1
ch+=1
if ch==2*(n-mul):
mul+=1
ch=0
nex=mul
if r==n*(n-1)+1:
print(1,end=" ")
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 1000000007
INF = float('inf')
# ------------------------------
def main():
def c(sm, a1):
sm = n*a1+(n-1)*n
for _ in range(N()):
n, l, r = RL()
i = n-1
sm = 0
while sm+i*2+1<l:
sm+=i*2 if i>0 else 1
i-=1
dif = l-sm
res = []
for j in range(n-i, n):
now = []
for k in range(j+1, n+1):
now.append(j)
now.append(k)
res+=now
# print(now)
if len(res)>(r-l+1)+dif: break
res = res[dif-1:]
if r==(n-1)*n+1: res.append(1)
# print(res)
print(*res[:r-l+1])
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 200010;
void solve(long long case_no) {
long long n, l, r, st = 0;
cin >> n >> l >> r;
vector<long long> res;
bool flag = false;
for (long long i = 1; i <= n; i++) {
if (st + res.size() >= r) break;
if (!flag and st + 2 * (n - i) < l) {
st += 2 * (n - i);
continue;
}
flag = true;
for (long long j = i + 1; j <= n; j++) res.push_back(i), res.push_back(j);
}
if (st + res.size() < r) res.push_back(1);
l -= st + 1;
r -= st + 1;
for (long long i = l; i <= r; i++) cout << res[i] << " ";
cout << '\n';
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long test_cnt = 1, case_no = 1;
cin >> test_cnt;
while (case_no <= test_cnt) solve(case_no++);
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class Euler {
public static void main(String[] args) throws IOException {
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(f.readLine());
PrintWriter out = new PrintWriter(System.out);
for (int t1 = 0; t1 < t; t1++) {
StringTokenizer tokenizer = new StringTokenizer(f.readLine());
int n = Integer.parseInt(tokenizer.nextToken());
long l = Long.parseLong(tokenizer.nextToken()) - 1;
long r = Long.parseLong(tokenizer.nextToken()) - 1;
int level = 1;
long index = 0;
boolean hadFirst = false;
while (level < n) {
long end = index + (n - level) * 2 - 1;
if (intersect(l, r, index, end)) {
for (int i = 0; i < n - level; i++) {
if (l <= index + i * 2 && index + i * 2 <= r) {
if (hadFirst) {
out.print(" ");
out.print(level);
} else {
out.print(level);
hadFirst = true;
}
}
if (l <= index + i * 2 + 1 && index + i * 2 + 1 <= r) {
if (hadFirst) {
out.print(" ");
out.print(level + i + 1);
} else {
out.print(level + i + 1);
hadFirst = true;
}
}
}
}
index = end + 1;
level++;
}
if (l <= index && index <= r) {
if (hadFirst) {
out.print(" ");
out.print(1);
} else {
out.print(1);
hadFirst = true;
}
}
out.println();
}
out.close();
}
private static boolean intersect(long start1, long end1, long start2, long end2) {
return Math.min(end1, end2) >= Math.max(start1, start2);
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long l, r;
int n;
vector<long long> a;
int main() {
int tt;
scanf("%d", &tt);
while (tt--) {
a.clear();
scanf("%d%lld%lld", &n, &l, &r);
long long sum = 0, pos1 = 0;
for (int i = 1; i <= n; i++) {
if (1 + sum <= l && 1ll * (n - i) * 2 + sum >= l) {
pos1 = i;
break;
} else
sum = sum + 1ll * (n - i) * 2;
}
if (l == 1ll * n * (n - 1) + 1)
a.push_back(1);
else {
while (l <= r) {
if (l % 2)
a.push_back(pos1);
else {
long long t = l - sum - 1;
t = pos1 + (t + 1) / 2;
a.push_back(t);
}
l++;
if (l > r) break;
if (l > (1ll * n - pos1) * 2 + sum) {
sum = sum + (1ll * n - pos1) * 2;
pos1++;
}
if (l == 1ll * n * (n - 1) + 1) {
a.push_back(1);
break;
}
}
}
int cnt = a.size();
for (int i = 0; i < cnt - 1; i++) printf("%lld ", a[i]);
printf("%lld\n", a[cnt - 1]);
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long int N = 100005;
void solve() {
long long int i, j, k, n, m, ans = 0, cnt = 0, sum = 0;
long long int l, r;
cin >> n >> l >> r;
vector<long long int> temp;
for (i = 1; i < n; i++) {
temp.push_back(2 * (n - i));
}
temp.push_back(1);
m = temp.size();
cnt = 1;
for (i = 0; i < m; i++) {
if (cnt + temp[i] > l) {
break;
} else {
cnt += temp[i];
}
}
i++;
if (i > n) {
i = 1;
}
j = i + 1;
ans = 0;
long long int cur = i;
while (cnt <= r) {
if (cnt == n * (n - 1) + 1) {
cout << 1;
break;
}
if (cnt >= l) {
cout << cur << " ";
}
if (ans == 0) {
cur = j;
j++;
} else {
if (j == n + 1) {
i++;
j = i + 1;
}
cur = i;
}
ans ^= 1;
cnt++;
}
cout << '\n';
return;
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int t;
cin >> t;
while (t--) solve();
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python2
|
import sys
__author__ = 'ratmir'
alphabet = "abcdefghijklmnopqrstuvwxyz"
def solve(n, a, graph):
return 1
def execute():
[t] = [int(x) for x in sys.stdin.readline().split()]
results = []
for ti in range(0, t):
#[n] = [int(x1) for x1 in sys.stdin.readline().split()]
[n, l, r] = [int(x1) for x1 in sys.stdin.readline().split()]
# ec = []
# for i in range(1, n+1):
# for j in range(i+1, n+1):
# ec.append(i)
# ec.append(j)
# ec.append(1)
cs = 1
cis = 0
while l>cis + 2 * (n-cs) and cs<n:
#print l, cis
cis += 2*(n-cs)
cs += 1
xx = l - cis - 1
arr = []
while r > cis and cs<n:
for i in range(cs+1, n+1):
arr.append(cs)
arr.append(i)
cis += 2*(n-cs)
cs += 1
arr.append(1)
#print xx
print(''.join('{} '.format(k) for k in arr[xx:xx+r-l+1]))
print(''.join('{}\n'.format(k) for k in results))
execute()
# 1 2 1
# 1 2 1 3 2 3 1
# 1 2 1 3 1 4 2 3 2 4 3 4 1
# 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 1
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
const int N = 100005;
void solve() {
long long n, s, e;
cin >> n >> s >> e;
if (s == n * n - n + 1) {
cout << 1 << '\n';
return;
} else {
long long a = n * 2 - 2;
long long sq = sqrt((a + 1) * (a + 1) - 4 * s);
long long st = (a + 1 - sq) / 2;
if (sq * sq < (a + 1) * (a + 1) - 4 * s) st++;
long long prev = (st - 1) * (a - (st - 2));
if (prev >= s) {
st--;
prev = (st - 1) * (a - (st - 2));
}
long long l = st, r = (s - prev + 1) / 2 + st;
if ((s % 2) == 0) {
cout << r << " ";
r++;
s++;
}
while (s <= min(e, n * n - n)) {
if (r > n) {
l++;
r = l + 1;
}
cout << l << " ";
s++;
if (s <= e)
cout << r << " ";
else
break;
r++;
s++;
}
if (e == n * n - n + 1) {
cout << 1 << '\n';
} else
cout << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
int t = 1;
cin >> t;
for (int i = 1; i <= t; i++) {
solve();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
python3
|
# -*- coding: utf-8 -*-
import os
import sys
from io import BytesIO, IOBase
INF = 2**62-1
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input():
return sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def slv(N, L, R):
k = 1
ans = []
for i in range(N-1):
c = (N-1-i)*2
if k + c >= L:
for j in range(c//2):
if L <= k <= R:
ans.append(i+1)
k += 1
if L <= k <= R:
ans.append(i+1+j+1)
k += 1
if k > R:
return ans
else:
k += c
if L <= k <= R:
ans.append(1)
return ans
def main():
T = read_int()
for _ in range(T):
N, L, R = read_int_n()
print(*slv(N, L, R))
if __name__ == '__main__':
main()
|
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