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codeforces round The Monster and the Squirrel 529B (数学规律)

时间:2023-04-20 17:38:12浏览次数:49  
标签:Monster Input codeforces she Ada Ari ray 529B first


The Monster and the Squirrel


Time Limit: 1000MS

 

Memory Limit: 262144KB

 

64bit IO Format: %I64d & %I64u


Submit Status


Description



Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feeding her squirrel.

Ari draws a regular convex polygon on the floor and numbers it's vertices 1, 2, ..., n in clockwise order. Then starting from the vertex 1 she draws a ray in the direction of each other vertex. The ray stops when it reaches a vertex or intersects with another ray drawn before. Ari repeats this process for vertex 2, 3, ..., n




Ada the squirrel wants to collect all the walnuts, but she is not allowed to step on the lines drawn by Ari. That means Ada have to perform a small jump if she wants to go from one region to another. Ada can jump from one region P to another region Q if and only if P and Q share a side or a corner.

Assuming that Ada starts from outside of the picture, what is the minimum number of jumps she has to perform in order to collect all the walnuts?


Input



The first and only line of the input contains a single integer n (3 ≤ n ≤ 54321) - the number of vertices of the regular polygon drawn by Ari.


Output



Print the minimum number of jumps Ada should make to collect all the walnuts. Note, that she doesn't need


Sample Input



Input



5



Output



9



Input



3



Output



1


Hint



One of the possible solutions for the first sample is shown on the picture above.

 

//题意可转化为给你一个凸多边形,每个点有标号为1,2,3,........n。从1向不与它相邻的点连线,然后依次是2,3,....n,但后边的不能穿过前面的线。问这么多的线把这个凸多边形分为几部分。

枚举前几个发现规律

#include<stdio.h>
#include<string.h>
int main()
{
	__int64 n,m;
	while(scanf("%I64d",&n)!=EOF)
	{
		printf("%I64d\n",(n-2)*(n-2));
	}
	return 0;
}

 

标签:Monster,Input,codeforces,she,Ada,Ari,ray,529B,first
From: https://blog.51cto.com/u_16079508/6210019

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