Codeforces Round #691 (Div. 2) B. Move and Turn 数学

最新推荐文章于 2024-05-04 09:45:46 发布

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最新推荐文章于 2024-05-04 09:45:46 发布

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本文探讨了一个机器人在无限二维平面上的行走策略。机器人从原点出发，每秒移动1米，且每次移动后必须改变方向90度。文章提供了一种算法来计算经过n步后，机器人可能达到的不同位置数量。

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A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.

The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.

Input
The only line contains a single integer n (1≤n≤1000) — the number of steps the robot makes.

Output
Print a single integer — the number of different possible locations after exactly n steps.

Examples
inputCopy
1
outputCopy
4
inputCopy
2
outputCopy
4
inputCopy
3
outputCopy
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.

In the second sample case, the robot will always end up 2–√ meters north-west, north-east, south-west, or south-east.
分奇偶看，如果是偶数那么就可以水平垂直走相同的数字，如果是奇数则有一个方向会多走一步。

#include<iostream>
#include<string>
#include<map>
#include<algorithm>
#include<memory.h>
#include<cmath>
#include<assert.h>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int a[maxn],b[maxn];
vector<int>l,r;
struct node{
	int l,r;
}n1[maxn];
void solve(){
	int n;
	cin>>n;
	if(n%2==1){
		int cnt =(n+1)/2;
		int ans=(cnt+1)*(cnt)/2;
		cout<<ans*4<<endl;
	}
	else{
		cout<<((n+2)/2)*((n+2)/2)<<endl;
	}
}
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	solve();
	return 0;
}