Problem Description
Dragon is watching NBA. He loves James and Miami Heat.
Here's an introduction of basketball game:http://en.wikipedia.org/wiki/Basketball. However the game in Dragon's version...
分类:
其他好文 时间:
2014-08-26 15:38:56
阅读次数:
183
Little Hasan loves to play number games with his friends.One day they were playing a game where one of them will speak out a positive numberand the others have to tell the sum of its factors. The firs...
分类:
其他好文 时间:
2014-08-25 11:57:04
阅读次数:
235
??ZCC Loves CodefiresTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 223Accepted Submission(s): 12...
分类:
其他好文 时间:
2014-08-23 13:53:30
阅读次数:
207
题目链接Jane loves string more than anything. She made a function related to the string some days ago and forgot about it. She is now confused about calcu...
分类:
其他好文 时间:
2014-08-22 19:36:09
阅读次数:
294
C. Petya and SpidersLittle Petya loves training spiders. Petya has a boardn?×?min size. Each cell of the board initially has a spider sitting on it. A...
分类:
其他好文 时间:
2014-08-17 01:04:31
阅读次数:
245
hdu 4878 ZCC loves words(AC自动机+dp+矩阵快速幂+中国剩余定理)
题意:给出若干个模式串,总长度不超过40,对于某一个字符串,它有一个价值,对于这个价值的计算方法是这样的,设初始价值为V=1,假如这个串能匹配第k个模式串,则V=V*prime[k]*(i+len[k]),其中prime[k]表示第k个素数,i表示匹配的结束位置,len[k]表示第k个模式串的长度(注...
分类:
其他好文 时间:
2014-08-14 20:37:19
阅读次数:
359
自从做了多校,整个人都不好了,老是被高中生就算了,题老是都不懂=-=原谅我是个菜鸟,原谅我智力不行。唯一的水题。Problem Description Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called .....
分类:
其他好文 时间:
2014-08-09 21:06:09
阅读次数:
259
题意:一个最大10^18*10^18的矩阵,给你最多十万个士兵的位置,分别分布在矩阵里,可能会位置重复,然后有2种操作,一种是把第i个士兵向上下左右移动,另一种是把第i个士兵与他横坐标纵坐标相同的士兵全部移到这个点上,然后要计算花费。
这道题我想了好几天。在看了标程得到一些提示后总算写出来了。加了读入优化后快了100ms左右达到546ms。
做法:开2个set分别维护X相同的和Y相同的,但是会...
分类:
其他好文 时间:
2014-08-06 19:20:22
阅读次数:
465
E. Lucky ArrayPetya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky...
分类:
其他好文 时间:
2014-08-02 23:22:24
阅读次数:
389
题目大意:求最少被扣除的时间
策略 如题;
对于两个邻近的题目i, j,对于他们对于在他们之前解决的题目的总时间的贡献t是不影响的,对于他们之后的总时间也不影响
这就推得对每一对相邻的他们对前后都是无影响的, 如果是交换的话原来是(t+e[i])*k[i] + (t+e[i]+e[j])*k[j], 就变成了(t+e[j])*k[j] + (t+e[i]+e[j])*k[i]
改变的就是...
分类:
其他好文 时间:
2014-07-31 17:13:26
阅读次数:
173