A11987 | Non-Coprime Partition
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内存限制256MB
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题目描述
Find out if it is possible to partition the first $n$ positive integers into two non-empty disjoint sets $S_1$ and $S_2$ such that:
$\mathrm{gcd}(\mathrm{sum}(S_1), \mathrm{sum}(S_2)) > 1$ Here $\mathrm{sum}(S)$ denotes the sum of all elements present in set $S$ and $\mathrm{gcd}$ means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from $1$ to $n$ should be present in exactly one of $S_1$ or $S_2$ .
$\mathrm{gcd}(\mathrm{sum}(S_1), \mathrm{sum}(S_2)) > 1$ Here $\mathrm{sum}(S)$ denotes the sum of all elements present in set $S$ and $\mathrm{gcd}$ means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
Every integer number from $1$ to $n$ should be present in exactly one of $S_1$ or $S_2$ .
输入格式
The only line of the input contains a single integer $n$ ( $1 \le n \le 45\,000$ )
输出格式
If such partition doesn't exist, print "No" (quotes for clarity).
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing $S_1$ and $S_2$ respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing $S_1$ and $S_2$ respectively.
Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.
If there are multiple possible partitions — print any of them.
输入输出样例
输入 #1
1
输出 #1
No
输入 #2
3
输出 #2
Yes 1 2 2 1 3
In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".
In the second example, the sums of the sets are $2$ and $4$ respectively. The $\mathrm{gcd}(2, 4) = 2 > 1$ , hence that is one of the possible answers.
In the second example, the sums of the sets are $2$ and $4$ respectively. The $\mathrm{gcd}(2, 4) = 2 > 1$ , hence that is one of the possible answers.
C++ 编辑器
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评测结果:Accepted