zoukankan      html  css  js  c++  java
  • CF div2 328 B

    B. The Monster and the Squirrel
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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 (in this particular order). And then she puts a walnut in each region inside the polygon.

    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 to leave the polygon after.

    Sample test(s)
    Input
    5
    Output
    9
    Input
    3
    Output
    1
    直接输出(n-2)的平方就好了,用long long
     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 #include <stack>
     5 #include <queue>
     6 #include <map>
     7 #include <algorithm>
     8 #include <vector>
     9 
    10 using namespace std;
    11 
    12 const int maxn = 1000005;
    13 
    14 typedef long long LL;
    15 
    16 vector<int>G[maxn];
    17 
    18 int a[maxn];
    19 int res[maxn];
    20 
    21 
    22 int main()
    23 {
    24   LL n;
    25   scanf("%lld",&n);
    26 
    27     printf("%lld
    ",(n-2)*(n-2));
    28 
    29 
    30     return 0;
    31 }
    View Code


  • 相关阅读:
    abap开发报表的简单过程
    有关innerHTML的知识
    MS SQL中的return&output的學習
    js获取下拉列表选中项的值和文本(select)以及获取单选按钮(radio)组的值和修改选中项[转]
    ASP.NET AJAX Control Toolkit
    VB.net下有个函数strconv可以进行简体繁体转换
    JavaScript:prototype属性使用说明
    [转]一份ASP内存的释放的实验报告
    什么是 Virtual Machine Additions(虚拟机附加安装模块)?
    Trigger&Procedure的應用
  • 原文地址:https://www.cnblogs.com/lmlyzxiao/p/4929532.html
Copyright © 2011-2022 走看看