D. Relatively Prime Graph
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Let's call an undirected graph G=(V,E)G=(V,E) relatively prime if and only if for each edge (v,u)∈E(v,u)∈E GCD(v,u)=1GCD(v,u)=1 (the greatest common divisor of vv and uu is 11). If there is no edge between some pair of vertices vv and uu then the value of GCD(v,u)GCD(v,u) doesn't matter. The vertices are numbered from 11 to |V||V|.
Construct a relatively prime graph with nn vertices and mm edges such that it is connected and it contains neither self-loops nor multiple edges.
If there exists no valid graph with the given number of vertices and edges then output "Impossible".
If there are multiple answers then print any of them.
Input
The only line contains two integers nn and mm (1≤n,m≤1051≤n,m≤105) — the number of vertices and the number of edges.
Output
If there exists no valid graph with the given number of vertices and edges then output "Impossible".
Otherwise print the answer in the following format:
The first line should contain the word "Possible".
The ii-th of the next mm lines should contain the ii-th edge (vi,ui)(vi,ui) of the resulting graph (1≤vi,ui≤n,vi≠ui1≤vi,ui≤n,vi≠ui). For each pair (v,u)(v,u) there can be no more pairs (v,u)(v,u) or (u,v)(u,v). The vertices are numbered from 11 to nn.
If there are multiple answers then print any of them.
Examples
input
Copy
5 6
output
Copy
Possible
2 5
3 2
5 1
3 4
4 1
5 4
input
Copy
6 12
output
Copy
Impossible
Note
Here is the representation of the graph from the first example:
题意：构造一个有n个顶点m条边的连通图，并且连通图相邻顶点互质。
思路：由连通图可知n个顶点最少有m-1（两个顶点一条边）条边，然后暴力即可。
#include<bits/stdc++.h>
using namespace std;
#define FOR(i,s,e) for(int i = (s); i < (e); i++)
#define FOE(i,s,e) for(int i = (s); i <= (e); i++)
#define FOD(i,s,e) for(int i = (s); i >= (e); i--)
#define ll long long
#define lb long double
#define pb push_back
#define mp make_pair
#define pff pair<double, double>
struct node
{
int x, y;
}edge[100005];
int main()
{
int n, m, t = 0, flag = 0;
cin >> n >> m;
if(n - 1 > m) { puts("Impossible"); ; return 0; }
FOE(i,1,n)
{
FOE(j,i+1,n)
{
if(__gcd(i,j) == 1)
{
edge[t].x = i;
edge[t++].y = j;
if(t == m)
{
flag = 1;
break;
}
}
}
if(flag == 1) break;
}
if(flag)
{
puts("Possible");
FOR(i,0,m) cout <<edge[i].x<<" "<<edge[i].y<<endl;
}
else puts("Impossible");
}