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# Yet Another Problem About Sequences CodeChef Solution

## Yet Another Problem About Sequences CodeChef Solution in C++17

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ll> vll;

#define all(v)       v.begin(),v.end()
#define sz(x)        (int)(x).size()
#define alr(v)       v.rbegin(),v.rend()
#define pb           push_back
#define S            second
#define F            first
#define pow2(x)      (1<<(x))
#define sp(x)        cout<<fixed<<setprecision(6)<<x
#define output(x)    cout<<(x?"YES\n":"NO\n")
#define bp(x)        __builtin_popcount(x)
//#define int long long
template<class T> using ordered_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// order_of_key (k) : Number of items strictly smaller than k .
// find_by_order(k) : K-th element in a set (counting from zero).

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

const int inf=2e9;
const int N=1e6+1;
const char nl='\n';
const ll mod=1e9+7;
const ll Binf=1e18;
const ll mod1=998244353;

vector<int> lp(N+1);
vector<int> pr;
void liner_sieve(){
for (int i=2; i <N; ++i) {
if (lp[i] == 0) {
lp[i] = i;
pr.push_back(i);
}
for (int j=0; j < (int)pr.size() && pr[j] <= lp[i] && i*pr[j] <= N; ++j) {
lp[i * pr[j]] = pr[j];
}
}
}

void solve101(){
// debug(sz(pr));
int n;
cin>>n;
int rem=n%3;

if(rem==0)
cout << "6 10 15 " ;
else if(rem==1)
cout << "21 14 10 15 ";
else
cout << "33 77 14 10 15 ";

n-=n%3+3;
vector<int>temp={6,10,15};
int j=0;
for(int i=0;i<n;i++){
cout<<pr[i+5]*temp[j]<<" ";
j=(j+1)%3;
}
cout<<nl;
}
signed main() {

ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);

int t=1;
liner_sieve();
// debug(pr);
cin>>t;
for(int i=1;i<=t;i++){
// cout<<"Case #"<<i<<": ";
solve101();
}
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
return 0;
}
// always check for binary search >.<

// #ifndef ONLINE_JUDGE
//     if (fopen("input.txt", "r"))
//     {
//         freopen("input.txt", "r", stdin);
//         freopen("output.txt", "w", stdout);
//     }
// #endif``````

## Yet Another Problem About Sequences CodeChef Solution in C++14

``````#include<bits/stdc++.h>
using namespace std;

#define ll long long int
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007

int main()
{
FIO;
int t,n,i,j,arr[]={6,10,15};
bool a[1000001]={0};

vector<int> v;
for(i=2;i<1000001;i++)
if(!a[i]){
v.push_back(i);
for(j=i;j<1000001;j+=i)
a[j]=true;
}

cin >> t;
while(t--)
{
cin >> n ;

i=n%3;

if(i==0)
cout << "6 10 15 " ;
else if(i==1)
cout << "21 14 10 15 ";
else
cout << "33 77 14 10 15 ";

n-=n%3+3;

for(i=0;i<n;i++)
cout << v[i+5]*arr[i%3] << ' ';

cout << '\n';
}
return 0;
}
//output``````

## Yet Another Problem About Sequences CodeChef Solution in PYTH 3

``````# cook your dish here
from sys import stdin,stdout
a=[0 for i in range(50005)]
pri=[0  for i in range(1000005)]
p=[0 for i in range(1000005)]

def prime(n):
ps=0
for i in range(2,n+1):
if p[i]==0:
pri[ps+1]=i
ps+=1
j=1
while j<=ps and i*pri[j]<=n:
p[i*pri[j]]=1
if i%pri[j]==0:
break
j+=1

def main():
t=int(t)
prime(1000000)
#print(pri[0:100])
for _ in range(t):
n=int(n)
if n==3:
stdout.write("6 10 15\n")
elif n==4:
stdout.write("10 15 21 14\n")
elif n==5:
stdout.write("78 10 105 77 143\n")
else:
cnt=(3-(n-1)%4)*4
j=3
for i in range(1,n+1,4):
if i<=cnt:
stdout.write(str(2*pri[j])+" "+str(pri[j]*pri[j+1])+" "+str(2*pri[j+1])+" ")
else:
stdout.write(str(2*pri[j])+" "+str(3*pri[j])+" "+str(3*pri[j+1])+" "+str(2*pri[j+1])+" ")
j+=2
stdout.write("\n")

main()``````

## Yet Another Problem About Sequences CodeChef Solution in C

``````#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#define mod 1000000007

long long int max=612000;
int primes[612000];

void sieve(int n)
{
//if 0 then prime else if 1 not prime
// int primes[max];
if(n<=3333)
{
max=200000;
}
else
{
max=612000;
}
long long int i,j;
memset(primes,0,max*(sizeof(primes[0])));
for(i=2;i<max;i++)
{
if(primes[i]==0)
{
for(j=2*i;j<max;j=j+i)
{
primes[j]=1;
}
}
}

}
void printsequence(int n)
{
sieve(n);
long long int b[n],k=0,i;
for(i=13;i<max;i++)
{
if(k<n)
{
if(primes[i]==0)
{
b[k]=i;
k++;
}
}
else
{
break;
}

}
//printf("k last=%lld\n",b[k-1]);
if(n==3)
{
printf("6 15 10\n");
}
else
{
long long int ans[n];
k=0;
for(i=0;i<(n-2);i++)
{
if(i%3==0)
{
ans[i]=b[k]*6;
}
else if(i%3==1)
{
ans[i]=b[k]*15;
}
else
{
ans[i]=b[k]*10;
}
k++;
}
k--;
//i--;
ans[i]=b[k]*7;
//printf("i==%lld\n",b[k]);
i++;
ans[i]=7*11;
ans[0]=ans[0]*11;
for(i=0;i<n;i++)
{
printf("%lld ",ans[i]);
}
printf("\n");
}
}
int main()
{
int m,t;
scanf("%d",&t);
for(m=0;m<t;m++)
{
int n;
scanf("%d",&n);
printsequence(n);
}
return 0;
}``````

## Yet Another Problem About Sequences CodeChef Solution in JAVA

``````/* package codechef; // don't place package name! */

import java.util.*;
import java.io.*;

class Sequences {
static PrintWriter out;
static final int MAXN = (int) 1e6;
static int spf[] = new int[MAXN];
static ArrayList<Integer> primes = new ArrayList<>();

public static void main(String[] args) throws Exception {
out = new PrintWriter(System.out);

sieveOfEratosthenes(MAXN);
int tc = in.nextInt();
for(int t = 0; t < tc; t++) {
solve();
}

out.flush();
out.close();
}

static void solve() throws Exception {
int n, seq[];
n = in.nextInt();
seq = new int[n];
Arrays.fill(seq, 1);
int resp[] = new int[]{2, 3, 5};
int np = 4, respi = 0;

for(int i = 0; i < n; i++) {
int mul;
if(i % 2 == 0) {
mul = primes.get(np++);
}
else {
mul = resp[respi];
respi = (i+1)%3;
}

if(i == n-1) mul = 7;

seq[i] *= mul;
seq[(i+1)%n] *= mul;
}

//        pn(isValid(seq, n));
for(int i = 0; i < n; i++) {
p(seq[i] + " ");
}
pn("");
}

static boolean isValid(int[] seq, int n) {
int pg;
for(int i = 0; i < n; i++) {
if((pg = gcd(seq[i], seq[(i+1)%n])) == 1 || gcd(pg, seq[(i+2)%n]) != 1) {
pn(seq[i] + " " + seq[(i+1)%n] + " " + seq[(i+2)%n]);
return false;
}
}

return true;
}

static void sieveOfEratosthenes(int n) {
boolean prime[] = new boolean[n+1];
for(int i=0;i<n;i++)
prime[i] = true;
for(int p = 2; p*p <= n; p++) {
if(prime[p]) {
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}

for(int i = 2; i <= n; i++) {
if(prime[i])
}
}

static int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b);}
static void p(Object o) {out.print(o);}
static void pn(Object o) {out.println(o);}
static void pnf(Object o) {out.println(o);out.flush();}

final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;

{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
}

{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
}

{
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;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)
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;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)
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;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)

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
{
buffer[0] = -1;
}

{
fillBuffer();
return buffer[bufferPointer++];
}

public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
}``````

## Yet Another Problem About Sequences CodeChef Solution in PYPY 3

``````def sieve():
n = 1000000
lp = [0]*(n+1)
pr = []

for i in range(2,n+1):
if lp[i] == 0:
lp[i] = i
pr.append(i)

j = 0
while(j < len(pr) and pr[j] <= lp[i] and (i * pr[j]) <= n):
lp[i*pr[j]] = pr[j]
j = j+1

return pr

o = sieve()
o.pop(0)
o.pop(0)
o.pop(0)
o.pop(0)
o.pop(0)
for _ in range(int(input())):
n = int(input())
if n == 3:
l = [6,10,15]
print(*l)
continue

if n == 4:
l = [21, 14, 10, 15]
print(*l)
continue

if n == 5:
l = [33, 77, 14, 10, 15]
print(*l)
continue

if n%3 == 0:
l = [6,10,15]
n = n-3
cnt = 0
while(cnt < n):
if cnt%3 == 0:
l.append(6*o[cnt])
#print(o[cnt])

if cnt%3 == 1:
l.append(10*o[cnt])
#print(o[cnt])

if cnt%3 == 2:
l.append(15*o[cnt])
#print(o[cnt])

cnt = cnt+1

print(*l)
continue

if n%3 == 1:
l = [21,14,10,15]
n = n-4
cnt = 0
while(cnt < n):
if cnt%3 == 0:
l.append(6*o[cnt])
#print(o[cnt])

if cnt%3 == 1:
l.append(10*o[cnt])
#print(o[cnt])

if cnt%3 == 2:
l.append(15*o[cnt])
#print(o[cnt])

cnt = cnt+1

print(*l)
continue

if n%3 == 2:
l = [33, 77, 14, 10, 15]
n = n-5
cnt = 0
while(cnt < n):
if cnt%3 == 0:
l.append(6*o[cnt])
#print(o[cnt])

if cnt%3 == 1:
l.append(10*o[cnt])
#print(o[cnt])

if cnt%3 == 2:
l.append(15*o[cnt])
#print(o[cnt])

cnt = cnt+1

print(*l)
continue``````

## Yet Another Problem About Sequences CodeChef Solution in PYTH

``````# cook your code here
from fractions import gcd
arr = []
def func(arr,n):
p=2
prime = [True for i in range(n+1)]
while(p*p<=n):
if prime[p]==True:
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
for p in range(2,n):
if prime[p]:
arr.append(p)

return arr
arr = func(arr,7*(10**5))

for _ in range(input()):
n = input()
ans = []
if n==3:
print 6,15,10
elif n==4:
print 374,595,1365,858
elif n==5:
print 6,15,35,77,286
elif n==7:
print 6,15,10,42,165,65,26
elif n==8:
print 6,15,10,42,165,130,13*19,19*2
elif n%3==0:
ans=[6,15,10]
for i in range(0,(n/3)-1):
ans.append(arr[(i+1)*3]*6)
ans.append(arr[(i+1)*3+1]*15)
ans.append(arr[(i+1)*3+2]*10)
print ' '.join(str(x) for x in ans)

elif n%3==1:
ans=[6,15,10]
for i in range(0,(n/3)-2):
ans.append(arr[(i+1)*3]*6)
ans.append(arr[(i+1)*3+1]*15)
ans.append(arr[(i+1)*3+2]*10)
ans.append(arr[(i+2)*3]*6)
ans.append(arr[(i+2)*3+1]*15)
ans.append(arr[(i+2)*3+2]*5)
ans.append(arr[(i+2)*3+2]*2)
print ' '.join(str(x) for x in ans)
elif n%3==2:
ans=[6,15,10]
for i in range(0,(n/3)-2):
ans.append(arr[(i+1)*3]*6)
ans.append(arr[(i+1)*3+1]*15)
ans.append(arr[(i+1)*3+2]*10)
ans.append(arr[(i+2)*3]*6)
ans.append(arr[(i+2)*3+1]*15)
ans.append(arr[(i+2)*3+2]*5)
ans.append(arr[(i+2)*3+2]*7)
ans.append(14)
print ' '.join(str(x) for x in ans)

``````

## Yet Another Problem About Sequences CodeChef Solution in C#

``````using System;
using System.Collections.Generic;
using System.Text;

namespace CodechefLongJan2019
{
class EARTSEQ
{
public static void Main()
{
Solve();
}

public static void Solve()
{
List<int> primes = seive(650000);
//Console.WriteLine(primes.Count);
int[] arr = new int[] {6, 10, 15};
while(t--> 0)
{
int remain = n % 3;
StringBuilder sb = new StringBuilder();
switch (remain)
{
case 0:
sb.Append("6 10 15");
break;
case 1:
sb.Append("21 14 10 15");
break;
case 2:
sb.Append("33 77 14 10 15");
break;

}
n = n - remain - 3;

for(int i = 0; i < n; i++)
{
sb.Append(" ");
sb.Append(arr[i % 3] * primes[i + 3]);
}
Console.WriteLine(sb.ToString());
}
}
/// <summary>
/// Calulate primes from 0 to n inclusive
/// </summary>
/// <param name="n"></param>
/// <returns>List of primes.</returns>
private static List<int> seive(int n)
{
List<int> prime = new List<int>();
bool[] nums = new bool[n + 1]; // marked if visited.
nums[0] = nums[1] = true;
for(int i = 2; i * i <= n; i++)
{
if(!nums[i])
{
for(int j = i * i; j <= n; j += i)
{
nums[j] = true;
}
}
}
for(int i = 2; i <= n; i++)
{
if(!nums[i])
{
}
}
return prime;
}
}
}
``````

## Yet Another Problem About Sequences CodeChef Solution in GO

``````package main

import (
"bufio"
"fmt"
"os"
)

var writer = bufio.NewWriter(os.Stdout)

func scanf(f string, a ...interface{})  { fmt.Fscanf(reader, f, a...) }
func scan(a ...interface{})             { fmt.Fscan(reader, a...) }
func printf(f string, a ...interface{}) { fmt.Fprintf(writer, f, a...) }
func print(a ...interface{})            { fmt.Fprint(writer, a...) }

// MAX var
var MAX = 700000

func main() {
defer writer.Flush()

primes := Sieve(MAX)
var n, t int
var r []int
scanf("%d\n", &t)
for ; t > 0; t-- {
scanf("%d\n", &n)
r = Solve(n, primes)
for i := 0; i < len(r); i++ {
printf("%d ", r[i])
}
printf("\n")
}
}

// Sieve func
func Sieve(n int) []int {
isprime := make([]bool, n)
prime := make([]int, 0)
SPF := make([]int, n)
for i := 2; i < n; i++ {
isprime[i] = true
}
for i := 2; i < n; i++ {
if isprime[i] {
prime = append(prime, i)
SPF[i] = i
}

for j := 0; j < len(prime) &&
i*prime[j] < n && prime[j] <= SPF[i]; j++ {
isprime[i*prime[j]] = false
SPF[i*prime[j]] = prime[j]
}
}
return prime
}

// Solve func
func Solve(n int, primes []int) []int {
r := make([]int, n)
r[0] = 2
r[1] = 2
for i := 1; i < n-1; i++ {
if i%3 == 0 {
r[i] *= 7
r[i+1] = 7
} else if i%3 == 1 {
r[i] *= 3
r[i+1] = 3
} else {
r[i] *= 5
r[i+1] = 5
}
}
r[n-1] *= 11
r[0] *= 11
for i := 0; i < n; i++ {
r[i] *= primes[i+5]
}
return r
}``````
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