C. Prime Swaps

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have an array a[1], a[2], ..., a[n], containing distinct integers from 1 to n. Your task is to sort this array in increasing order with the following operation (you may need to apply it multiple times):

• choose two indexes, i and j (1 ≤ i < j ≤ n; (j - i + 1) is a prime number);
• swap the elements on positions i and j; in other words, you are allowed to apply the following sequence of assignments:tmp = a[i], a[i] = a[j], a[j] = tmp (tmp is a temporary variable).

You do not need to minimize the number of used operations. However, you need to make sure that there are at most 5n operations.

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains n distinct integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).

Output

In the first line, print integer k (0 ≤ k ≤ 5n) — the number of used operations. Next, print the operations. Each operation must be printed as "i j" (1 ≤ i < j ≤ n; (j - i + 1) is a prime).

If there are multiple answers, you can print any of them.

Examples

input

3
3 2 1

output

1
1 3

input

2
1 2

output

0

input

4
4 2 3 1

output

3
2 4
1 2
2 4

  HERE WE GO THROUGH A GREEDY STRATEGY AND ALWAYS BRING THE NUMBER TO ITS NEAREST POSITION THAT CAN BE DONE USING A VALID SWAP. IT IS ALWAYS CORRECT BECAUSE WE CAN SWAP TWO NEIGHBORING NUMBERS ALWAYS SO THIS WILL LEAD TO A SOLUTION.

            LOOK AT THE EASY CODE

#include<bits/stdc++.h>

using namespace std;

#define ll long long int 

bool pr[100010];

#define ff first

#define ss second

#define mp make_pair

#define pb push_back

vector<int> v;

int l;

void pre()

{

   for(int i=2;i<=100000;i++)

   {

      if(!pr[i])

      {

        v.pb(i);

        for(int j=2*i;j<=100000;j+=i)

        {

              pr[j]=1;

  }

  }

}

l=v.size();

}

int pos[100010];

int a[100010];

int main()

{

    pre();

     int n;

     cin>>n;

     for(int i=1;i<=n;i++)

     {

       cin>>a[i];

       pos[a[i]]=i;

 }

 vector<pair<int,int> > res;

 

 for(int i=1;i<=n;i++)

 {

        while(pos[i]!=i)

        {

          int lst=pos[i];

       

             int shiftm=abs(pos[i]-i)+1;

   vector<int> ::iterator it;

   it=upper_bound(v.begin(),v.end(),shiftm);

it--;

int val=*it;

// cout<<"value "<<val<<endl;

     int idx=pos[i]+1-val;

//      cout<<"idx "<<idx<<endl;

     res.pb(mp(pos[i],idx));

     

     pos[a[idx]]=pos[i];

     pos[i]=idx;

     

    // cout<<"sert pos of "<<i<<" as "<<pos[i]<<" and pos of "<<a[idx]<<" = "<<pos[a[idx]]<<endl;

     swap(a[idx],a[lst]);

}

 

}

int sz=res.size();

cout<<sz<<endl;

for(int i=0;i<sz;i++)

cout<<min(res[i].ff,res[i].ss)<<" "<<max(res[i].ff,res[i].ss)<<endl;

 return 0;

 }

All submissions for this problem are available.

Read problems statements in Mandarin Chinese and Russian.

You are given a array A of N positive integers, and you can perform the following operation on the array

1) Pick any two indices i and j in the array

2) Divide A[i] and A[j] by some common factor of A[i] and A[j]



You can perform the above operation as many number of times as you want, and the aim is to minimize the product of the resulting array. Find this minimum product. Since the answer can be a large number, print the product modulo 1000000007 (10⁹+7).



INPUT:

First line contains T, number of testcases. Each testcase contains 2 lines.

First line of each testcase contains single integer N, size of the array

Second line of each testcase contains N space separated integers, the array A



OUTPUT:

For each testcase, output single line indicating the answer to that testcase



CONSTRAINTS:

1<=T<=10

30 points : 1<=N<=2000, 1<=A[i]<=10⁶

70 points : 1<=N<=20000, 1<=A[i]<=10⁸



SAMPLE INPUT:

1

3

2 3 6



SAMPLE OUTPUT:

1



EXPLANATION:

First divide first and third numbers by 2, then the second and third by 3. This makes all numbers equal to 1, hence the product is 1.

CODE IS AS FOLLOWS

//Coder: Balajiganapathi
#define TRACE
#define DEBUG

#include <algorithm>
#include <bitset>
#include <deque>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <utility>
#include <vector>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<string> vs;

// Basic macros
#define st          first
#define se          second
#define all(x)      (x).begin(), (x).end()
#define ini(a, v)   memset(a, v, sizeof(a))
#define re(i,s,n)   for(int i=s;i<(n);++i)
#define rep(i,s,n)  for(int i=s;i<=(n);++i)
#define fr(i,n)     re(i,0,n)
#define repv(i,f,t) for(int i = f; i >= t; --i)
#define rev(i,f,t)  repv(i,f - 1,t)
#define frv(i,n)    rev(i,n,0)
#define pu          push_back
#define mp          make_pair
#define sz(x)       (int)(x.size())

const int oo = 2000000009;
const double eps = 1e-9;

#ifdef TRACE
    #define trace1(x)                cerr << #x << ": " << x << endl;
    #define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
    #define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
    #define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
    #define trace5(a, b, c, d, e)    cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
    #define trace6(a, b, c, d, e, f) cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;

#else

    #define trace1(x)
    #define trace2(x, y)
    #define trace3(x, y, z)
    #define trace4(a, b, c, d)
    #define trace5(a, b, c, d, e)
    #define trace6(a, b, c, d, e, f)

#endif

const int mx = 20004, mod = 1000000007;
int a[mx], n;

set<int> primes;
map<int, vector<pi> > pr;  // Key is prime and the value is a vector of number of times the prime divides a[i]

//Decompose each number into its prime factors and populate the above map
void getPrimes() {
    primes.clear();
    pr.clear();
    fr(i, n) {
        int m = a[i];
        for(int p = 2; p <= m / p; ++p) {
            int cnt = 0;
            while(m % p == 0) {
                ++cnt;
                m /= p;
            }
            if(cnt > 0) {
                primes.insert(p);
                pr[p].pu(mp(cnt, i));
                //trace3(i, p, cnt);
            }
        }
        if(m > 1) {
            primes.insert(m);
            pr[m].pu(mp(1, i));
        }
    }
}

int main() {
    int t;
    cin >> t;
    assert(1 <= t && t <= 10);

    while(t--) {
        cin >> n;
        assert(1 <= n && n <= 20000);
        fr(i, n) cin >> a[i];
        fr(i, n) assert(1 <= a[i] && a[i] <= 100000000);
        getPrimes();
        vi v(all(primes));

        //Solve for each prime separately using a greedy strategy: Take the two elements with the largest no. of the prime currently and divide them by the prime. Decrement the count and update them.
        fr(j, sz(v)) {
            int p = v[j];
            //trace1(p);
            priority_queue<pi> q;
            fr(j, sz(pr[p])) {
                q.push(pr[p][j]);
                //trace3(j, pr[p][j].st, pr[p][j].se);
            }
            while(sz(q) > 1) { //Atleast two elements needed
                pi p1 = q.top(); q.pop();
                pi p2 = q.top(); q.pop();
                a[p1.se] /= p;
                a[p2.se] /= p;
                if(p1.st > 1) q.push(mp(p1.st - 1, p1.se));
                if(p2.st > 1) q.push(mp(p2.st - 1, p2.se));
            }
        }

        ll ans = 1;
        fr(i, n) {
            ans *= a[i];
            ans %= mod;
        }
        cout << ans << endl;

    }
 return 0;
}

Larry's Array

locked

by nikasvanidze

• Problem
• Submissions
• Leaderboard
• Discussions
• Editorial

Larry has a permutation of N$N$ numbers, A$A$, whose unique elements range from 1$1$ to N$N$ (i.e.: A={a1,a2,…,aN−1,aN}$A=\{a_1,a_2,\dots ,a_{N-1},a_N\}$). He wants A$A$ to be sorted, so he delegates the task of doing so to his robot. The robot can perform the following operation as many times as it wants:

• Choose any 3$3$ consecutive indices and rotate their elements in such a way that ABC$ABC$rotates to BCA$BCA$, which rotates to CAB$CAB$, which rotates back to ABC$ABC$.

For example: if A={1,6,5,2,4,3}$A=\{1,6,5,2,4,3\}$ and the robot rotates (6,5,2)$(6,5,2)$, A$A$ becomes {1,5, 2, 6,4,3}$\{1,\mathbf{\text{5, 2, 6}},4,3\}$.

On a new line for each test case, print YES$\mathtt{\text{YES}}$ if the robot can fully sort A$A$; otherwise, print NO$\mathtt{\text{NO}}$.

Input Format

The first line contains an integer, T$T$, the number of test cases. 
The 2T$2T$ subsequent lines each describe a test case over 2$2$ lines:

1. An integer, N$N$, denoting the size of A$A$.
2. N$N$ space-separated integers describing A$A$, where the ith$i^{th}$ value describes element ai$a_i$.

Constraints

• 1≤T≤10$1\le T\le 10$
• 3≤N≤1000$3\le N\le 1000$
• 1≤ai≤N, where every element ai is unique.$1\le a_i\le N\text{, where every element }{a}_i\text{ is unique.}$

Output Format

On a new line for each test case, print YES$\mathtt{\text{YES}}$ if the robot can fully sort A$A$; otherwise, print NO$\mathtt{\text{NO}}$.

Sample Input

3
3
3 1 2
4
1 3 4 2
5
1 2 3 5 4

Sample Output

YES
YES
NO

Let's call inversion in sequence A$A$ (size of N$N$) number of pairs(x,y)$(x,y)$,
where 1≤x<y≤N$1\le x<y\le N$ and Ax>Ay$A_x>A_y$.


Let's prove that parity of inversions will not change:
A,B,C$A,B,C$ -> B,C,A$B,C,A$.
Order of A,C$A,C$ changed thats ±1$\pm 1$.
Order of A,B$A,B$ changed thats ±1$\pm 1$.
So change is −2$-2$ or 0$0$ or 2$2$.

Prove that we can always get permutation:
[1,2,3,...,n−2,n−1,n]$[1,2,3,...,n-2,n-1,n]$ or [1,2,3,...,n−2,n,n−1]$[1,2,3,...,n-2,n,n-1]$. (n$n$ and n−1$n-1$ are swapped). using out operation we are moving B$B$ and C$C$ left by one, in this way we can move 1$1$ in first place, then move 2$2$ to second place and so on including n−2$n-2$.
So, number of inversions will be 0([1,2,3,...,n−2,n−1,n])$0([1,2,3,...,n-2,n-1,n])$ or 1([1,2,3,...,n−2,n,n−1])$1([1,2,3,...,n-2,n,n-1])$.
If it is 1$1$ then it's impossible to sort, and if it's 0 that means it's sorted.


Calculate parity using any BST, solution will be O(T∗Nlog2N)$O(T\ast Nlo{g}_{2}N)$.

In this problem you can calculate answer by compairing each pair. O(T∗N2)













#include<bits/stdc++.h>

using namespace std;

int a[1010];

int main()

{

    int n;

    int t;

    cin>>t;

    while(t--)

    {

        

         cin>>n;

         for(int i=1;i<=n;i++)

         cin>>a[i];

         int in=0;

         for(int i=1;i<=n;i++)

         {

            for(int j=i+1;j<=n;j++)

      {

           if(a[j]<a[i])

        in++; 

      } 

   }

   if(in&1)

   cout<<"NO"<<endl;

   else cout<<"YES"<<endl;

    }

    return 0;



}