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题意　：给定一个长度为n的区间．然后给k次询问，每次一个数t,求一个区间[l,r]使得这个区间和的绝对值最接近t

Jessica’s Reading Problem **Time Limit:** 1000MS **Memory Limit:** 65536K **Total Submissions:** 8787 **Accepted:** 2824 Description Jessica’s a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that text book, like other authors, is extremely fussy about the ideas, thus some ideas are covered more than once. Jessica think if she managed to read each idea at least once, she can pass the exam. She decides to read only one contiguous part of the book which contains all ideas covered by the entire book. And of course, the sub-book should be as thin as possible.

由于顺序是可以改变的． 所以考虑是否可以映射．只要存在字母对应出现的次数都相同．那么就可以通过映射得到． 具体是开一个数组记录每个字母出现的次数… 然后sort

Rabbit and Grass # **Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3058 Accepted Submission(s): 2261 **

hdu 2149题目链接 题意＆思路：巴什博奕，点m是n点。。。然后往前画即可。。。

题目链接：hdu 2188题目链接 题意＆思路：巴什博奕。。画n点p点。。。

**序：**博弈是信息学和数学试题中常会出现的一种类型，算法灵活多变是其最大特点，而其中有一类试题更是完全无法用常见的博弈树来进行解答。 寻找必败态即为针对此类试题给出一种解题思路。

给６个矩形的长和宽（或者宽和长），问这六个矩形能否组成一个长方体．

比赛的时候没过．还以为是树状数组写残了． 但实际上是有自己不知道的东西． 这种博弈叫　nim游戏 所以这是一个二维的nim游戏． **nim游戏的性质是xor 和为0必败，否则必胜． xor和也有前缀和性质，所以可以用树状数组维护．

假设有n堆石子，每堆石子的个数分别如下 a1, a2, a3, … an 定义nim-sum为a1^a2^a3…an

果然dp還是弱項啊啊啊啊．． 不過比最開始的完全無從下手強了不少應該．．．

贪心．．尽量把一样的材料放在一起．．． 然后写蠢了．．妈蛋．．． 详情见代码

x的二进制表示中1的个数即为答案． 原因是，每天晚上糖果数量翻倍，相当于左移１位,这时候二进制表示中1的数量不变

初识分快. 引一段题解： Let’s split rectangle 106 × 106 by vertical lines into 1000 rectangles 103 × 106. Let’s number them from left to right. We’re going to pass through points rectangle by rectangle. Inside the rectangle we’re going to pass the points in increasing order of y-coordinate if the number of rectangle is even and in decreasing if it’s odd.

因为每一个正整数可以唯一分解质因数… 要看能猜多少次，只要知道不大于n的质因子数有多少个即可（相同的算多

背包还是理解的不够透彻．． 因为每次都是用那个一维形式的． 这道题的做法类似01背包.