Search is not available for this dataset
name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.*;
import java.util.*;
public class MainClass
{
public static void main(String[] args)throws IOException
{
Reader in = new Reader();
int t = in.nextInt();
StringBuilder stringBuilder = new StringBuilder();
while (t-- > 0)
{
int n = in.nextInt()... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated meth... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | //package cf;
import java.io.*;
import java.util.*;
public class T_class {
static int m=2147483647;///this is 2^32 -1 (is prime number) and should be used for hashing
static int p=1000000007;
static int max=(int)2e3+10;
static boolean prime[]=new boolean[max+5];
static int dp[][]=new int[(int)max]... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for __ in range(int(input())):
n = int(input())
# a = [-1]*(n+1)
b = [-1]*(2*n)
a = [int(x) for x in input().split()]
m1 = {}
for i in range(n):
m1[a[i]] = 1
b[2*i] = a[i]
for i in range(1,2*n,2):
k = a[i//2]
while k in m1:
k += 1
b[i] = k
m1[k]=1
#print(m1)
f=0
for x in b:
if x > 2*n:
f=1... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int tc, ca = 0;
cin >> tc;
while (tc--) {
int n, el;
vector<int> b;
vector<int> baki;
map<int, int> m1;
map<int, int> m2;
cin >> n;
for (int i = 0; i < ... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 |
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;
public class realfast implements Runnable {
private static final int INF = (int) 1e9;
long in= (long)Math.pow(10,9);
public void solve() throws IOException
{
int t ... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import sys
input = sys.stdin.readline
t = int(input())
import bisect
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
b = [0 for _ in range(2*n)]
for x in a:
b[x-1] = 1
c = []
for i in range(2*n):
if b[i] == 0:
c.append(i+1)
try:
d = []
for x in a:
bi = bi... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 |
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.TreeSet;
/**
* @author Mubtasim Shahriar
*/
public class ResPer ... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for _ in range(int(input())):
n = int(input())
l = [*map(int,input().split())]
s = set()
s1 = set(l)
for i in range(1,2*n + 1):
if(i not in s1):
s.add(i)
ans = []
for i in range(n):
ans.append(l[i])
for j in range(l[i] + 1, 2 * n + 1,1):
if(j i... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long int fast_expo(long long int x, long long int p) {
if (p == 0)
return 1;
else if (p % 2 == 0) {
long long int t = fast_expo(x, p / 2) % 1000000007;
return (t * t) % 1000000007;
} else
return (x * (fast_expo(x, p - 1)) % 1000000007) % 100000000... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t=int(input())
for _ in range(t):
n=int(input())
b=list(map(int, input().split() ))
s=set(b)
ans=[]
possible=True
for x in b:
y=x+1
while y<=2*n:
if y not in s: break
y=y+1
if y>2*n:
possible=False
break
s.add(y)
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.*;
import java.util.*;
public class MyClass {
public static void main(String args[]) {
FastReader sc = new FastReader(); //For Fast IO
//func f = new func(); //Call func for swap , permute, upper bound and for sort.
StringBuilder sb = new StringBuilder();
... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for nt in range(int(input())):
n=int(input())
b=list(map(int,input().split()))
d=set(b)
ansflag=0
ans=[]
d1={}
for i in range(n):
ans.append(b[i])
k=1
flag=0
#print (ans)
while True:
if (b[i]+k)>2*n:
flag=1
break
if b[i]+k not in d and b[i]+k not in d1:
ans.append(b[i]+k)
d1[b[i]+... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.math.*;
public class Codeforces {
public static void main(String[] args) throws IOException {
Scanner s = new Scanner(System.in);
int t=s.nextInt();
for(int i=0;i<t... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int arr[n];
unordered_set<int> us;
for (int i = 0; i < n; i++) {
cin >> arr[i];
us.insert... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read integers
get_int = lambda: int(sys.stdin.readline())
#to print fast
#pt = lambda x: sys.stdout.write(str(x)+'\n')
#------... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for _ in range(int(input())):
n = int(input())
arr = list(map(int,input().split()))
ans = [0 for i in range(2*n)]
visited = [False for i in range(2*n)]
ind = 0
for i in range(0,2*n,2):
ans[i] = arr[ind]
visited[arr[ind]-1]=True
ind+=1
# print(visited)
for i in ran... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | whatever=int(input())
for i in range (whatever) :
k=int(input())
bs=list(map(int,input().split()))
a=[0]*2*k
avail=[]
for t in range (1,(2*k)+1) :
if t not in bs :
avail.append(t)
avail=sorted(avail)
for l in range (0,(2*k),2) :
a[l]=bs[int(l/2)]
flag=0 ... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int b[105], vis[205];
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n;
scanf("%d", &n);
int pd = 0;
memset(vis, 0, sizeof(vis));
int maxx = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &b[i]);
maxx = max(maxx, b[i]);
... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import bisect
import collections
t = int(input())
for test in range(t):
canDo = True
n = int(input()) # elements in sequence
b = []
a = list(range(1,2*n+1))
for i in input().split():
b.append(int(i))
a.remove(int(i))
a.sort()
c = []
for i, elem in enumerate(b):
p... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for _ in " "*int(input()):
a=int(input())
if a==1:
z=int(input())
if z==1:print("1 2")
else:print(-1)
continue
b=list(map(int,input().split()))
if 1 not in b or 2*a in b:print(-1);continue
c=[ i for i in range(1,(2*a)+1) if i not in b]
r=[]
for i in range(a):
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.util.*;
import java.io.*;
import java.math.*;
public class Main7
{
static class Pair
{
int x;
int y;
int z;
Pair(int x, int y,int z)
{
this.x=x;
this.y=y;
this.z=z;
}
}
static int mod=1000000007;
public static int[] sort(int[] a)
{
int n=a.length;
ArrayList<Integer> ar=new Arr... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<long long> vprime;
void SieveOfEratosthenes(int n) {
bool prime[n + 1];
memset(prime, true, sizeof(prime));
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p) prime[i] = false;
}
}
for (int p = 2; p ... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.*;
public class ProblemC {
public static InputStream inputStream = System.in;
public static OutputStream outputStream = System.out;
public static void main(String[] args) {
Scanner scanner = new... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t=int(input())
for _ in range(t):
n=int(input())
b=list(map(int,input().split()))
s=set(range(2 * n + 1))
s.remove(0)
l=[201]*(2*n)
flag = True
for i in range(0, 2*n, 2):
l[i] = b[i//2]
s.remove(l[i])
for i in range(1, 2*n, 2):
for num in s:
if num > l... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
const int N = 1e6;
const double eps = 0.00000001;
const long long mod = 1e9 + 7;
using namespace std;
long long read() {
long long x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x ... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, a[100], b[100], i, j, f;
cin >> t;
while (t--) {
cin >> n;
int temp[201] = {0};
for (i = 0; i < n; i++) {
cin >> a[i];
temp[a[i]]++;
}
for (i = 0; i < n; i++) {
f = 0;
j = a[i] + 1;
while (j <= 2... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.*;
import java.util.ArrayList;
import java.util.List;
public class Solution {
public static void main(String[] args) throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(System.out);
... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t=int(input())
s=[]
for njj in range(0,t):
n=int(input())
b=list(map(int,input().split( )))
if 1 not in b or 2*n in b:
s.append([-1])
else:
a=[]
for i in b:
g=1
f=i
while i+g in a or i+g in b :
g+=1
if i+g>2*n:
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) {
long long n;
cin >> n;
long long arr[n];
map<long long, long long> mp;
for (long long i = 0; i < n; i++) {
cin >> arr[i];
mp[arr[... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
cin >> n;
vector<int> v(2 * n + 1, 0), A;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
v[x] = 1;
A.push_back(x);
}
int justice = 0;
for (int i = 1; i <= 2 * n; i++) {
if (justice < 0 && v[i] == 1) {
cout << -1 ... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | from sys import exit
from bisect import bisect
def iis(): return map(int, input().split())
def ii(): return int(input())
def liis(): return list(map(int, input().split()))
t = ii()
for _ in range(t):
n = ii()
b = liis()
todos = []
duplas = []
for i in range(1, 2*n+1):
if i not in b:
todos.append(i)
flag = ... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | from collections import *
for i in range(int(input())):
n, a, ans = int(input()), list(map(int, input().split())), []
mem = Counter(a)
for j in range(n):
ans.append(a[j])
for k in range(a[j] + 1, 2 * n + 1):
if not mem[k]:
ans.append(k)
mem[k] = ... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
b = list(map(int, input().split()))
ansls=[0]*(2*n)
used=set(b)
flag=True
for i in range(n):
ansls[2*i]=b[i]
for j in range(b[i]+1, 2*n+1):
if not j in used:
ansls[2*i+1]=j
used.add(j... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
for x in range(t):
n = int(input())
b = list(map(int, input().split()))
numbers_used = [False for i in range(2*n)]
a = []
for i in b:
numbers_used[i-1] = True
for i in b:
a.append(i)
for j in range(i, 2*n):
if numbers_used[j]==False:
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for _ in range(int(input())):
n, b = int(input()), list(map(int, input().split()))
d = {e: True for e in range(1, 2*n + 1)}
a = [0] * (2 * n)
for i in range(n):
a[2*i] = b[i]
d[b[i]] = False
ok = True
for i in range(1, 2*n, 2):
e = a[i-1] + 1
while e <= 2*n:
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | n = int(input())
for i in range(n):
lena = int(input())
arr = list(map(int, input().split()))
rem = [j for j in list(range(1, 2*lena+1)) if j not in arr]
counter = 0
final = []
for j in range(lena):
counter2 = 0
for k in range(arr[j], 2*lena+1):
if k in rem:
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
b = list(map(int, input().split()))
for i in range(n):
b[i] -= 1
# print("b: ", b)
used = [False for _ in range(2*n)]
for i in range(n):
used[b[i]] = True
ans = [0 for i in range(2*n)]
for i in range(n):
ans[2*i] = b[i]
result = True
for i in range(n... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 |
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
b = set(range(1,n*2+1))
b = b - set(a)
b = list(b)
f = 0
ans = []
for i in a:
found = 0
j = 0
while j < len(b):
if b[j] > i:
found = 1
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 100 + 10;
int n;
int b[maxn];
int ans[2 * maxn];
void solve() {
cin >> n;
map<int, int> m;
m.clear();
for (int i = 1; i <= n; i++) {
cin >> b[i];
m[b[i]] = 1;
ans[2 * i - 1] = b[i];
}
for (int i = 1; i <= ... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t=int(input())
for _ in range(t):
n=int(input())
a=[i for i in range(1,2*n+1)]
m=[int(i) for i in input().split()]
l=[]
for e in m:
k=e
l.append(k)
while(True):
e=e+1
if e not in m and e not in l:
l.append(e)
break
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for _ in range(int(input())):
n = int(input())
b = list(map(int,input().split()))
used = [False]*( (2*n)+1)
a = [None]*(2*n)
possible=True
for bi in b: used[bi] = True
for i in range(2*n):
if i%2==0:
a[i] = b[i//2]
else:
no = a[i-1]+1
while... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
bool u = 0, n = 0, z = 0;
int a, b, c[100000] = {}, p, d, q = 1, s = 1;
cin >> a;
for (int i = 0; i < a; i++) {
cin >> b;
p = b * 2;
s = 1;
for (int i = 0; i <= (b * 2 - 2); i += 2) cin >> c[i];
for (int i = 0; i <= (b * 2 - 2); i +=... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
for test in range(t):
n = int(input())
b = [int(i) for i in input().split()]
def solve():
def minimal(v):
for i in perm:
if i > v:
return perm.pop(perm.index(i))
return v
perm = list(range(1, n * 2 + 1))
for... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Comparator;
import java.util.Scanner;
import java.util.Arrays;
import java.util.TreeSet;
import java.util.StringTokenizer;
public class vk18
{
public static void main(String[] stp) throws Exception
{
... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public cla... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 |
def STR(): return list(input())
def INT(): return int(input())
def MAP(): return map(int, input().split())
def MAP2():return map(float,input().split())
def LIST(): return list(map(int, input().split()))
def STRING(): return input()
import string
import sys
from heapq import heappop , heappush
from bisect import *
from... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
solutions = []
for i in range(t) :
n = int(input())
string = str(input())
st = string.split()
lst = []
for j in range(n) :
lst.append(int(st[j]))
solution = []
for j in range(n) :
solution.append(lst[j])
number = lst[j]
ok = 0
while ok... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for i in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = []
d = 0
for i in range(1, 2*n+1):
if i not in a:
b.append(i)
b.sort()
c = []
i = 0
j = 0
while(len(b) > 0):
if(b[-1] < a[i]):
d = 1
break
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | I=input
for _ in[0]*int(I()):
n=2*int(I());a=n*[0];b=a[::2]=*map(int,I().split()),;c={*range(1,n+1)}-{*b};i=1
try:
for x in b:z=a[i]=min(y for y in c if x<y);c-={z};i+=2
except:a=-1,
print(*a) | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | from math import *
# from sys import stdin,stdout
def binarySearch(arr,x,i):
l=i
r=len(arr)-1
while l <= r:
mid = (l + r)//2;
if arr[mid] == x:
return mid
elif arr[mid] < x:
l = mid + 1
else:
r = mid - 1
return -1
def js(arr,x):
l=0
r=len(arr)-1
ans=-1
while(l<=r):
m=(l+r)//2
if(ar... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.util.*;
import java.io.*;
public class Asd {
static Scanner s=new Scanner(System.in);
static PrintWriter w=new PrintWriter(System.out);
public static void main(String args[])
{
int t=s.nextInt();
while(t-->0)
{
solve();
}
w.close();
}
public static void... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.util.*;
public class Solution{
public static void main(String args[]){
Scanner in=new Scanner (System.in);
int t=in.nextInt();
while(--t>=0){
int n=in.nextInt();
int a[]=new int[n];
int b[]=new int[2*n];
HashSet<Integer> hset=new Ha... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
/*
HashMap<> map=new HashMap<>();
TreeMap<> map=new TreeMap<>();
map.put(p,map.getOrDefault(p,0)+1);
for(Map.Entry<> mx:map.entrySet()){
int v=mx.getValue(),k=mx.getKey();
}for (int i = 1; i <= 1000; i++)
... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | 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.util.InputMismatchException;
import java.io.IOException;
import java.util.HashSet;
import java.io.Writer;
import java.io.OutputStreamWr... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
M: while (t-- > 0) {
int n = scn.nextInt(), B_ARR[] = new int[n + 1];
HashSet<Integer> set = new HashSet<>();
for (int i = 1; i <= n; i++) {
B_ARR... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.*;
import java.util.*;
import java.math.*;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class RestoringPermutation {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputRe... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
import heapq
for _ in range(t):
n = int(input())
B = list(map(int, input().split()))
if 1 not in set(B):
print(-1)
continue
d = {}
A = [0]*(2*n)
for i in range(n):
A[2*i] = B[i]
d[B[i]] = 2*i+1
#print(id)
flag = True
ids = []
heapq... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import math
from collections import Counter
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
ll=[]
s=list(range(1,n+1))
for i in l:
ll.append(i)
if i+1 not in ll and i+1 not in l:
ll.append(i+1)
else:
j=i+2
whil... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | from bisect import *
for t in range(int(input())):
n=int(input())
b=list(map(int,input().split()))
f=0
c=set(b)
d=[]
for i in range(1,2*n+1):
if i not in c:
d.append(i)
a=[]
for i in range(n):
a.append(b[i])
x=bisect_left(d,b[i])
if x==len(d):
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.util.*;
import java.io.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
InputReader in = new InputReader(System.in);
PrintWriter out = new PrintWriter(System.out);
int t= in.nextInt();
while(t-- > 0) {
... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | for _ in range(int(input())):
n = int(input())
dic = {}
for i in range(1,2*n+1):
dic[i] = 0
arr = list(map(int,input().split()))
flag = 0
for i in range(n):
if arr[i]>=2*n:
flag = 1
dic[arr[i]] = 1
l = [0]*(2*n)
count = 0
for i in range(0,2*n,2):
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import sys
import copy
input=sys.stdin.buffer.readline
t=int(input())
while t:
t-=1
n=int(input())
arr=list(map(int,input().split()))
arr1=copy.deepcopy(arr)
arr1.sort()
if arr1[0]==1:
if max(arr)+1<=2*n:
ans=[]
for i in range(n):
ans.append(arr[i])
for j in range(arr[i]+1,2*n+1):
if j not i... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int t, n;
void solve() {
cin >> n;
int b[n];
bool foundMin = false;
bool foundMax = false;
for (int i = 0; i < n; i++) {
cin >> b[i];
if (b[i] == 1) foundMin = true;
if (b[i] == n * 2) foundMax = true;
}
if (!foundMin || foundMax) {
cout << "-1... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool compare(const pair<string, string>& i, const pair<string, string>& j) {
if (i.first != j.first)
return i.first < j.first;
else
return i.second < j.second;
}
vector<long long> factorize(long long n) {
vector<long long> v;
for (long long i = 2; i * i <= n... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[101];
bool c[201];
vector<int> b[101];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
memset(c, false, sizeof(c));
for (int i = 0; i < n; i++) {
b[i].clear... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.*;
import java.util.*;
public class Main {
static Scanner sc = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
static void solve() throws Exception {
int n = sc.nextInt(), arr[] = new int[2 * n];
TreeSet<Integer> st = new TreeSet<>();
for (int i = 1; i <= 2 * n;... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | t = int(input())
for _ in range(t):
n = int(input())
B = list(map(int, input().split()))
cands = [True] * (2 * n)
for b in B:
cands[b - 1] = False
ret = []
for b in B:
ret.append(b)
for i in range(2 * n):
if b < i + 1 and cands[i]:
ret.append(i + 1)
cands[i] = False
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int t, i, j, k, n, x, y, x1, y1, x2, y2, p, arr[1000], b[2000], l;
cin >> t;
while (t--) {
cin >> n;
j = 0;
long long int flag = 0, flag1 = 1;
for (i = 0; i < n; i++) {
cin >> arr[i];
if (arr[i] > n) {
j++;
... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | T=int(input())
for _ in range(T):
a=int(input())
A=list(map(int,input().split()))
DP=[0 for i in range((2*a)+1)]
flag=0
for i in A:
if(i>(2*a)):
flag=1
else:
DP[i]=1
if(flag==1):
print(-1)
continue
else:
mv=a
B=[]
... | PYTHON3 |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int arr[105];
unordered_set<int> used;
bool fl = 1;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> arr[i];
used.insert(arr[i]);
}
vector<int> v;
for (int i = 0; i < n; ++i) {
... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int b[n];
int ans[2 * n];
for (long long i = 0; i < n; i++) cin >> b[i];
set<int> a;
for (long long i = 0; i < 2 * n; ... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
public class CodeForce {
public static void main(String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(Sy... | JAVA |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
cin >> n;
int arr[n];
vector<bool> brr(2 * n + 1, 0);
brr[0] = 1;
for (int i = 0; i < n; i++) {
cin >> arr[i];
brr[arr[i]] = 1;
}
vector<int> v(n);
bool flag = 1, flag2 = 0;
for (int i = 0; i < n; i++) {
flag2 = 0;
f... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
set<int> s;
for (int i = 1; i <= 2 * n; i++) s.insert(i);
vector<int> b(n + 1);
vector<int> a(2 * n + 1, 0);
... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n), b(2 * n);
multiset<int> m;
for (size_t i = 0; i < 2 * n; ++i) m.insert(i + 1);
for (a... | CPP |
1315_C. Restoring Permutation | You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).
The first line of... | 2 | 9 | def bubbleSort(arr):
n = len(arr)
# Traverse through all array elements
for i in range(n):
# Last i elements are already in place
for j in range(0, n-i-1):
# traverse the array from 0 to n-i-1
# Swap if the element found is greater
# than ... | PYTHON3 |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
using LL = long long;
namespace _buff {
const size_t BUFF = 1 << 19;
char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
char getc() {
if (ib == ie) {
ib = ibuf;
ie = ibuf + fread(ibuf, 1, BUFF, stdin);
}
return ib == ie ? -1 : *ib++;
}
} // namespace _buff
LL read() {
... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
namespace IO {
static const int IN_BUF = 1 << 23, OUT_BUF = 1 << 23;
inline char myGetchar() {
static char buf[IN_BUF], *ps = buf, *pt = buf;
if (ps == pt) {
ps = buf, pt = buf + fread(buf, 1, IN_BUF, stdin);
}
return ps == pt ? EOF : *ps++;
}
template <typename T>
inline bool read(... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
using i64 = long long;
const int maxN = 223456;
const int P = 998244353;
int n, m, rnk;
i64 base[maxN], p[maxN], dp[60][60][2];
int cnt[maxN], ans[maxN];
void dfs1(int d, i64 x) {
if (d == rnk) {
cnt[__builtin_popcountll(x)]++;
} else {
dfs1(d + 1, x);
dfs1(... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int N = 200010;
const int M = 60;
inline int add(int u, int v) { return (u += v) >= MOD ? u - MOD : u; }
inline int sub(int u, int v) { return (u -= v) < 0 ? u + MOD : u; }
inline int mul(int u, int v) { return (long long)u * v % MOD; }
inli... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
template <class T>
inline void read(T &x) {
x = 0;
register char c = getchar();
register bool f = 0;
while (!isdigit(c)) f ^= c == '-', c = getchar();
while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
if (f) x = -x;
}
const int N = 60, mod = 998244353;
int n, m, k, t, c[N];
lo... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
namespace {
using u64 = uint64_t;
static const int MOD = 998244353;
struct LargeSolver {
LargeSolver(int r_, int m_, std::vector<u64> &&b_)
: r{r_}, m{m_}, b{std::move(b_)}, count(m + 1), result(m + 1) {
dfs_large(0, 0, 0);
int power_of_two = 1;
for (int i = 0; i < m - r; ++... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
#pragma GCC target("popcnt,sse4.2")
using namespace std;
const int mod = 998244353;
inline int add(int a, int b) {
if ((a += b) >= mod) a -= mod;
return a;
}
inline int dec(int a, int b) {
if ((a -= b) < 0) a += mod;
return a;
}
inline int mult(int a, int b) {
long long t = 1ll * a * ... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
void exgcd(int a, int b, int& x, int& y) {
if (!b) {
x = 1, y = 0;
} else {
exgcd(b, a % b, y, x);
y -= (a / b) * x;
}
}
int inv(int a, int n) {
int x, y;
exgcd(a, n, x, y);
return (x < 0) ? (x + n) : (x);
}
const int Mod = 998244353;
template <const... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int mo = 998244353;
int n, m, s1, cnt[65536];
long long a[55], x, ans[55];
void insert(long long x) {
for (int i = (int)(m - 1); i >= (int)(0); i--)
if (x & (1ll << i))
if (!a[i]) {
a[i] = x;
break;
} else
x ^= a[i];
}
int Cou... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
inline void add(int &x, int y) { (x += y) >= mod ? x -= mod : 0; }
inline int pl(int x, int y) { return (x += y) >= mod ? x - mod : x; }
inline int kpow(int a, int b) {
int s = 1;
for (; b; b >>= 1, a = 1ll * a * a % mod)
if (b & 1) s = 1l... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
long long mul(long long a, long long b) {
if (!b) return 1;
long long k = mul(a, b / 2);
if (b & 1) return k * k % 998244353 * a % 998244353;
return k * k % 998244353;
}
int n, m, k;
long long b[64], arr[64], C[64][64], w[64][64], ans[64];
long long val;
bool mark[6... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int mn = 1e5 + 10;
const int mm = 53;
const long long mod = 998244353;
long long basis[mm];
int m;
bool add(long long x) {
for (int i = m - 1; i >= 0; i--)
if ((x >> i) & 1) {
if (basis[i])
x ^= basis[i];
else {
basis[i] = x;
... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m;
vector<long long> bs;
int cnt[(1 << 16) + 1];
long long inv2 = (998244353 + 1) / 2;
int popcount(long long x) {
return cnt[x & 65535] + cnt[x >> 16 & 65535] + cnt[x >> 32 & 65535] +
cnt[x >> 48 & 65535];
}
bool add(long long x... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
template <class T>
inline void read(T &res) {
res = 0;
bool bo = 0;
char c;
while (((c = getchar()) < '0' || c > '9') && c != '-')
;
if (c == '-')
bo = 1;
else
res = c - 48;
while ((c = getchar()) >= '0' && c <= '9')
res = (res << 3) + (res << 1) + (c - 48);
if (... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int P = 998244353, N = 100;
int n, m, g[N], k, C[N][N], w[N][N], ans[N], cnt;
long long a[N], b[N], c[N], x;
int ksm(int a, int b) {
int ans = 1;
for (; b; b >>= 1, a = 1ll * a * a % P)
if (b & 1) ans = 1ll * ans * a % P;
return ans;
}
void insert(long long ... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class t, class u>
void chmax(t& first, u second) {
if (first < second) first = second;
}
template <class t, class u>
void chmin(t& first, u second) {
if (second < first) first = second;
}
template <class t>
using vc = vector<t>;
template ... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
for (; !isdigit(ch); ch = getchar()) {
if (ch == '-') f = -1;
}
for (; isdigit(ch); ch = getchar()) {
x = x * 10 + ch - 48;
}
return x * f;
}
const int mxN = 1 << 19;
const int mxM = 53;
const ... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
template <class T>
inline void chkmax(T &a, T b) {
if (a < b) a = b;
}
template <class T>
inline void chkmin(T &a, T b) {
if (a > b) a = b;
}
inline int read() {
int s = 0, f = 1;
char ch = getchar();
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '... | CPP |
1336_E2. Chiori and Doll Picking (hard version) | This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m;
vector<long long> bs, todo;
int cnt[(1 << 16) + 1];
long long inv2 = (mod + 1) / 2;
int popcount(long long x) { return __builtin_popcountll(x); }
long long getp(long long x) { return 63 - __builtin_clzll(x); }
bool add(long long x) {
f... | CPP |
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