POJ 2886 Who Gets the Most Candies?（反素数+数学推导+模拟+线段树||树状数组+二分）

N children are sitting in a circle to play a game.

The children are numbered from 1 to N in clockwise order. Each of them has a card with a non-zero integer on it in his/her hand. The game starts from the K-th child, who tells all the others the integer on his card and jumps out of the circle. The integer on his card tells the next child to jump out. Let A denote the integer. If Ais positive, the next child will be the A-th child to the left. If A is negative, the next child will be the (−A)-th child to the right.

The game lasts until all children have jumped out of the circle. During the game, the p-th child jumping out will get F(p) candies where F(p) is the number of positive integers that perfectly divide p. Who gets the most candies?

Input

There are several test cases in the input. Each test case starts with two integers  N(0 <  N  ≤ 500,000) and K (1 ≤ K ≤ N) on the first line. The next N lines contains the names of the children (consisting of at most 10 letters) and the integers (non-zero with magnitudes within 10⁸) on their cards in increasing order of the children’s numbers, a name and an integer separated by a single space in a line with no leading or trailing spaces.

Output

Output one line for each test case containing the name of the luckiest child and the number of candies he/she gets. If ties occur, always choose the child who jumps out of the circle first.

Sample Input

4 2
Tom 2
Jack 4
Mary -1
Sam 1

Sample Output

Sam 3

题解：

题意：

有n个人标号1,2...n围成圈，每个人有一张卡，有数字，开始从第k个人开始，他退出圈子，然后如果他的卡是正数x，那么从他开始向右数x个人为下一个出圈的人，如果是负数那么从他开始向左数x个人那个人出圈，求这一整个过程中，假设第i个人是第x次出圈，那么f（x）为x的因子个数，求f（x）最大值和该人的名字

线段树或者树状数组对于这题只是个工具。。重要的根据数学推导来模拟就很有难度了。。。。

对于这题求因子个数用筛法打个表就好了，线段树或者树状数组方面也很好写，但是对于数学推导+模拟我完全没啥思路。。。还是看了别人的博客理解了以后打了一遍，为什么线段树你要和数学联合起来欺负我这个弱渣。。。

我的代码里面因为看见数据范围是500000有点大不是很适合建树，有是单点更新，我就直接用树状数组了

我看的博客：http://www.cnblogs.com/ZefengYao/p/6617350.html

ps：关于求因子不懂的我引用上面的博客的一段话解释

首先需要制定一份因子表factor_num[N]，用于记录每一个数的因子的个数.做法类似于埃氏筛选。从小到大找素数，每找到一个素数，这个素数都有可能是比这个素数大的那些数的因子，每找到一个比当前素数大的并且能整除该素数的数，判断这个数中包含了多少当前素数因子。（108=3*3*3*2*2，所以它的素数因子为{1，3，9，27}*{1，2，4}，一共有4*3个因子）

我的代码：

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