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codeforces 1205A Almost Equal

2019-08-19 16:01:04  阅读:117  来源: 互联网

标签:1205A YES Almost codeforces int numbers output 2n example



You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:

For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.

For example, choose n=3. On the left you can see an example of a valid arrangement: 1+4+5=10, 4+5+2=11, 5+2+3=10, 2+3+6=11, 3+6+1=10, 6+1+4=11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5+1+6=12, and 3+2+4=9, 9 and 12 differ more than by 1.
在这里插入图片描述
Input
The first and the only line contain one integer n (1≤n≤105).

Output
If there is no solution, output “NO” in the first line.

If there is a solution, output “YES” in the first line. In the second line output 2n numbers — numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.

Examples
input

3

output

YES
1 4 5 2 3 6

input

4

output

NO

Note

Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.

题意:

给你一个n,我们需要构造一个长度为2n 的环,数字分别为1 ~ 2n,使环上每n个连续的点的记录下来。其中这些数要求只有2个数字构成,并且这两个数必须连续。没有为NO, 有的则输出YES 和方案。

题解:

每个点会在n个环上,所以所有环的总和为sum = (1 + 2 + … + 2 * n) * n.
一共有2n个环所以两个不同和的环的和(有点绕)为rsum = sum / n, 即为rsum = (1 + 2 + … 2 * n).
所以rsum 是偶数时,则无法分为两个连续的自然数。无解。
当rsum是奇数时,有解。
模拟填即可。
怎么填?
我们可以这样:

1

2
然后构造
1的左边从上往下填,1的右边从下往上填。
右下填3,
依次左边填2个,右边填2个(最后一次一个),填满即可。

代码如下:
#include <iostream>
#include <cstring>
using namespace std;
int a[200005];
int main(){
	int n;
	scanf("%d", &n);
	int m = 2 * n;
	long long r = (1 + m) * m / 2;
	if(r % 2 == 0){
		printf("NO\n");
		return 0;
	}
	printf("YES\n");
	a[1] = 1;
	a[n + 1] = 2;
	int q = 1, h = n + 2;
	a[n + 2] = 3;
	a[2 * n] = 2 * n;
	int now = 4;
	while(q + 2 <= n || h + 2 < 2 * n){
		if(q + 2 <= n){
			a[++q] = now++;
			a[++q] = now++;
		}
		if(h + 2 < 2 * n){
			a[++h] = now++;
			a[++h] = now++;
		}
	}
	for(int i = 1; i < 2 * n; i++){
		printf("%d ", a[i]);
	}
	printf("%d\n", a[2 * n]); 
	return 0; 
} 


标签:1205A,YES,Almost,codeforces,int,numbers,output,2n,example

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