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D. Prime Graph

分类: IT文章 • 2022-04-01 08:32:48

D. Prime Graph

Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!

When building the graph, he needs four conditions to be satisfied:

  • It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
  • The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
  • The total number of edges must be prime.
  • The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.

Below is an example for n=4.

D. Prime Graph

Note that the graph can be disconnected.

Please help Bob to find any such graph!

Input

The input consists of a single integer 3≤n≤1000) — the number of vertices.

Output

If there is no graph satisfying the conditions, print a single line containing the integer −1.

Otherwise, first print a line containing a prime number vi. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.

If there are multiple solutions, you may print any of them.

Note that the graph can be disconnected.

Examples
input
Copy
4
output
Copy
5
1 2
1 3
2 3
2 4
3 4
input
Copy
8
output
Copy
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note

The first example was described in the statement.

In the second example, the degrees of vertices are 13, is also a prime number, hence both conditions are satisfied.

D. Prime Graph
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod =1e9+7;
const int N = 1e5+5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}      

int main()
{
    int m,n;
    cin >> n;
    m = n;
    while (!prime(m))
        m++;
    cout << m << "
1 " << n << endl;
    rep(i, 1, n - 1)
        cout << i << " " << i + 1 << endl;
    rep(i, 1, m - n)
        cout << i << " " << i + n/2<< endl;
    return 0;
}
 

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