“玲珑杯”ACM比赛 Round #4 F -- My-graph【图的同构（构造）】

“玲珑杯”ACM比赛 Round #4

Start Time：2016-11-05 12:00:00 End Time：2016-11-05 17:00:00 Refresh Time：2016-11-06 22:00:19 Private

F -- My-graph

Time Limit：1s Memory Limit：64MByte

Submissions：359Solved：178

DESCRIPTION

In graph theory, the inverse of a graph $G$ is a graph $H$ on the same vertices such that two distinct vertices of $H$ are adjacent if and only if they are not adjacent in $G$.
My-graph is the graph that it's inverse graph is an isomorphism graph of itself.

(This is a example of My-graph has 5 nodes).
Now comes the problem: Do you have a way to make a My-graph that has $n$ nodes?

INPUT

There are multiple test cases.The first line is a number T ( $T\le 100$), which means the number of cases.For each case, one integer $n(2\le n\le {10}^4)$ means the nodes you have initially.

OUTPUT

If you have at least one way to make it , please output "yes".If there is no way, output "no".

SAMPLE INPUT

225

SAMPLE OUTPUT

noyes

SOLUTION

“玲珑杯”ACM比赛 Round #4

  图的同构看书上没有具体方法判断，给出结论吧：对于任意一个有 k 个顶点的且与自己互补的图，在它外面添加一根由 4 个新顶点组成的链条 x – y – z – w ，再把顶点 x 与所有 k 个顶点都相连，把 w 也与所有 k 个顶点都相连。容易看出，整个图现在就有 k + 4 个顶点了，并且这个新的图也和自己是互补的。

AC代码:

#include

typedef long long LL;

int main()
{
	int T; scanf("%d",&T);
	while(T--) {
	    int N; scanf("%d",&N);
		LL ans=N*(N-1)>>1;
		printf((ans&1)?"no":"yes");
		puts("");
	}
	return 0;
}