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[BZOJ1228][SDOI2009]E&D

2014-11-25 19:00:14| 分类： Problems | 标签： |举报 |字号大中小订阅

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1228: [SDOI2009]E&D

找了一节课的规律终于搞出来了

代码真是异常短

题解在代码下方

#include<cstdio>
#include<cstdlib>
int sg(int i,int j){
	unsigned long long tmp=2;
	for(int re=0;;re++,tmp*=2)
	if((i-1)%tmp<tmp/2 && (j-1)%tmp<tmp/2)return re;
}
int n;
int scan(){int i=0;scanf("%d",&i);return i;}
int main(){
	int i,j,a,b,tmp;
	int nn = scan();
	while(nn--){
		n = scan()/2;
		tmp=0;
		for(i=1;i<=n;i++){
			a = scan();b = scan();
			tmp ^= sg(a,b);
		}
		if(tmp)printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

题解：

SG函数

首先给sg函数打个表……发现0的分布好有规律

然后又发现1好像也挺规则的，一堆小三角型

最后发现每个数都是类似的三角形

发现如下规律：


	0出现条件
	i,j均%2=1
	1出现条件
	i%4=1,2
	j%4=1,2
	2出现条件
	i,j%8=1,2,3,4
	……

得到sg(i,j)=k的必要条件：

(i-1)%2^k+1 < 2^k

(j-1)%2^k+1 < 2^k

^{但是需要注意这里是必要不充分条件，因为每个数字并不是网格状，而是三角形。}

^{为什么呢，因为对于3，它某个合适的位置可能同时也对2合适，那么这个值就是2而不是3}

^{也就是说最sg(i,j)会等于最小的合适的k}

^i,j≤10^{9,所以直接暴力从小到大枚举，每次时间是logn}

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