DodgeBall - Hackerrank


http://www.1point3acres.com/bbs/thread-180481-1-1.html
 从1到n选择尽可能多的数,使得任意两个数不为倍数关系,问你最大能取多少个数(或者如题目所说的,至少多少个必然存在冲突)

第一反应也是找primes,结果举了个1~10的例子检验就发现不对了。。。

把1-n数字列出来,有整除关系的数之间加一条边,问题就转换成为找最大的集合使得集合里的数互相没有边,答案就是集合里数的个数+1,枚举几个n之后,我觉的答案就是n/2+1,因为n是偶数,所以[n/2 + 1, n]之间不能互相整除,n/2之前的数必然能整除[n/2 + 1, n]之间的某个数,因为将其乘2得到的结果必然在[n / 2 + 1, n]直接,所以我认为答案应该就是n/2 + 1


我觉得证明不难啊,用contradiction证,假设大于(n+1)/2,那么肯定有一些数落在(0,n/2),假设有x个,那么这些数的2倍就不可以出现在(n/2,n)中,所以(n/2,n)至多有n/2-x个,所以总数至多有(n+1)/2个,得出矛盾。

大概是这个意思,边界我就没细分析了

so 直接return n/2+1(感觉是这个),那还写啥...就像n lines divide plane那个题,直接return 1+(1+n)*n/2不就可以...But那个lz跑test case跑不过...?

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