Let's call an undirected graph?$\mathrm{G=(V,E)G=(V,E)?relatively\; prime?if\; and\; only\; if\; for\; each\; edge?(v,u)\in E(v,u)\in E??GCD(v,u)=1GCD(v,u)=1?(the\; greatest\; common\; divisor\; of?vv?and?uu?is?11).\; If\; there\; is\; no\; edge\; between\; some\; pair\; of\; vertices?vv?and?uu?then\; the\; value\; of?GCD(v,u)GCD(v,u)?doesn\text{'}t\; matter.\; The\; vertices\; are\; numbered\; from?11?to?|V||V|.}$

Construct a?relatively prime?graph with?$\mathrm{nn?vertices\; and?mm?edges\; such\; that\; it\; is\; connected\; and\; it\; contains\; neither\; self-loops\; nor\; multiple\; edges.}$

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?$\mathrm{nn?and?mm?(1\le n,m\le 1051\le n,m\le 105)\; —\; the\; number\; of\; vertices\; and\; the\; number\; of\; edges.}$

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?$\mathrm{ii-th\; of\; the\; next?mm?lines\; should\; contain\; the?ii-th\; edge?(vi,ui)(vi,ui)?of\; the\; resulting\; graph\; (1\le vi,ui\le n,vi\ne ui1\le vi,ui\le n,vi\ne ui).\; For\; each\; pair?(v,u)(v,u)there\; can\; be\; no\; more\; pairs?(v,u)(v,u)?or?(u,v)(u,v).\; The\; vertices\; are\; numbered\; from?11?to?nn.}$

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:

?

? ?這題無腦暴力 暴力真的出了奇跡?

? ?暴力枚舉一遍就行了

?

? ??

1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 10; 4 const int INF = 0x3fffffff; 5 typedef long long LL; 6 using namespace std; 7 int n, m; 8 struct node { 9 int x, y; 10 node () {} 11 node (int x, int y): x(x), y(y) {} 12 } qu[maxn]; 13 int main() { 14 scanf("%d%d", &n, &m); 15 if (n - 1 > m) { 16 printf("Impossible\n"); 17 return 0; 18 } 19 int k = 0, flag = 0; 20 for (int i = 1 ; i <= n ; i++) { 21 for (int j = i + 1 ; j <= n ; j++) { 22 if (__gcd(i, j) == 1) qu[k++] = node(i, j); 23 if (k == m) { 24 flag = 1; 25 break; 26 } 27 } 28 if (flag) break; 29 } 30 if (flag) { 31 printf("Possible\n"); 32 for (int i = 0 ; i < k ; i++) 33 printf("%d %d\n", qu[i].x, qu[i].y); 34 } else printf("Impossible\n"); 35 return 0; 36 }

?

轉載于:https://www.cnblogs.com/qldabiaoge/p/9314221.html

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