题目来源:http://poj.org/problem?id=1016 题目大意: 对一个非负整数定义一种运算(inventory):数这个数中各个数字出现的次数,然后按顺序记录下来.比如“5553141”有2个1,1个3,一个4,3个5,于是运算后为“21131435”,对于这种运算有的数字有着有趣的性质:如“31123314”,它运算后的结果和它本身是一样的(self-inventorying).如果做j次inventory运算后,第j次迭代的结果是self-inventorying的,则称…

点击打开链接 Numbers That Count Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17922   Accepted: 5940 Description "Kronecker's Knumbers" is a little company that manufactures plastic digits for use in signs (theater marquees, gas station…

Description        "Kronecker's Knumbers" is a little company that manufactures plastic digits for use in signs (theater marquees, gas station price displays, and so on). The owner and sole employee, Klyde Kronecker, keeps track of how many digi…

"Kronecker's Knumbers" is a little company that manufactures plastic digits for use in signs (theater marquees, gas station price displays, and so on). The owner and sole employee, Klyde Kronecker, keeps track of how many digits of each type he…

题意是将一串数字转换成另一种形式.比如5553141转换成2个1,1个3,1个4,3个5,即21131435.1000000000000转换成12011.数字的个数是可能超过9个的.n个m,m是从小到大排序. 输出的结果又四种情况,建议判断的时候就按照题目的顺序进行判断,否则可能出错.第一种情况是进行一次变换后和原来的数字相同:第二种是进行j次后和原来数字相同:第三种是变换过程中形成了循环,比如70:       70是2循环,又最后3组数字可判断出:第四种情况是超过15次变换仍然找不出规律.…

Tamref love random numbers, but he hates recurrent relations, Tamref thinks that mainstream random generators like the linear congruent generator suck. That's why he decided to invent his own random generator. As any reasonable competitive programmer…

Tamref love random numbers, but he hates recurrent relations, Tamref thinks that mainstream random generators like the linear congruent generator suck. That's why he decided to invent his own random generator. As any reasonable competitive programmer…

对于Java程序员来说,null是令人头痛的东西.时常会受到空指针异常 (NPE)的骚扰.连Java的发明者都承认这是他的一项巨大失误.Java为什么要保留null呢?null出现有一段时间了,并且我认为Java发明 者知道null与它解决的问题相比带来了更多的麻烦,但是null仍然陪伴着Java. 我越发感到惊奇,因为java的设计原理是为了简化事情,那就是为什么没有浪费时间 在指针.操作符重载.多继承实现的原因,null却与此正好相反.好吧,我真的不知道这个问题的答案,我知道的是不管null…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…

Fun one.. the punch line of this problem is quite common in Bit related problems on HackerRank - visualize it in your mind, and you will find: all bits on the same index among all numbers, will not involve other bits on other indices. So, we simply c…