其他

有的時候總會去想如何才能用更少的時間複雜度來解決問題，但是有的時候 ACM需要快速的敲出代碼，要利用暴力得到一個簡潔的解決方案。A ： http://codeforces.com/contest/252/problem/A暴力int data[110];int main(){int n;cin>>n;for(int i = 1 ; i <= n ; i ++ ) cin>>data[i];int maxn = -1;int x;for(int i = 1 ; i

The flood fill algorithm is somewhat tricky to implement. I use a iterative method instead of recursive flooding. I first scan through the west and east points to the current point until a different colored one is encountered. Then, the points

I convert all the calculations from floating-point to integral to avoid all the messes in floating-point calculation. The method is straight forward: I simulate the exchanging progress and use a 'budget' to record the extra cents that should be

 Let f(n) denote the number of ways to make the sum of n. When n > 3, the number of digits is greater than 1. The first digit d used must be one of 1, 2, 3 or 4. Remove the first digit, then the remaining sum is n-d. There are f(n-d) ways to make

Let n denotes the length of the expression, d denotes the depth at most, and the count is f(n,d). Obviously, f(n,d) = 0 when n is odd. For other conditions, let us consider where we can find the corresponding parenthesis of the leftmost one. It can

Let f(k, d) denote the number of labelings of the k-ary tree of depth d. Moreover, let n(k, d) denote the number of nodes of the tree. The tree contains k subtrees, each contains n(k, d-1) nodes. There are totally  (n(k, d) - 1)! / (n(k, d-1)!)^k

I develop a script to post my code to the UVa online judgeautomatically. This tool will also replace the local #include's withits real content when uploading the code to the server. So, I haven't to paste the code all around any more.Curl and

I believe sorting the poker at first may be helpful. You should determine the kind of the poker hand from the highest rank down to the lowest. Comparing the poker hands with the same kind is tricky to implement due to the complex rules.Code:/********

Let f(n) denote the furthest distance we can go with n steps, then we can derive this formula easily:   f(n) = ((n+1)/2)^2 if n is odd, (1, 2, 3, 4, 3, ,2, 1)    f(n) = (n/2)^2 + n/2 if n is even. (1, 2, 3, 3, 2, 1)Give the distance of f(n), we can

一個geohash的 值是對應的一個地圖地區的，然後我需要的這個地區的座標區間 ，即經緯度的最大最小值。那麼在geohash 的wiki首頁上找到c的實現，贊一個。在我的機器上還不能直接編譯通過，做了如下改變：1 geohash.c 檔案改成geohash.cpp2 去掉 stdbool.h 這是C99 的標頭檔，vs2005 不支援，stdbool的全文如下，實現不行，就賦值內容到geohash.h檔案中去#ifndef _STDBOOL_H#define _STDBOOL_H#ifndef

之前下載ubuntu 下的deb ，安裝會出現問題，一些包的依賴關係沒有得到滿足，我所使用的是wxWidgets 2.8 太老了 。之前在win7 上試用了12.11 ，感覺很好，代碼補全，和調試功能都有了很大的改進。首先貼一下，之前直接安裝之後，codeblocks 雖然可以用，但是之後你要再安裝其他軟體 ，有一個和wxWidgets 相關的 報錯，正在讀取軟體包列表... 完成正在分析軟體包的相依樹狀結構正在讀取狀態資訊... 完成vim 已經是最新的版本了。您可能需要運行“apt-get

 First, convert each package p = (s1, s2, s3) to a vector (s2 - s1, s3 - s1). All the packages forms convex hull. If point (0, 0) is in this convex hull, then there will be a solution, otherwise not. I can not tell the proof of this now, if you have

 The final state of the last bulb is determined by the number of factors of n. Only the number which is the square of a number has an odd number of factors, thus it has a state of "yes".Code:/**********************************************************

Output related problem. I use a predefined bitmap of all the numbers and enlarge them when displaying.Code:/************************************************************************* * Copyright (C) 2008 by liukaipeng                                  

From the description of this problem we can easily draw the recurrence of   f(n) = 2*f(n-k) + 2^k - 1and the base cases of   f(0) = 0 and f(1) = 1However, how can we determine which k would give the minimum value of f(n)?Suppose k is giving the

Compute the remainder with the following equation:    (x*y)%n = (x%n)*(y%n)%nCode:/*************************************************************************** *   Copyright (C) 2008 by Liu Kaipeng                                     * *   LiuKaipeng 

The close form of fibs does not help in the problem. The input range is too large that the precision of long double won't be sufficient to handle it correctly. Just generate all the fibs up to 100 digits and use string compare to find the

For the detail explanation of the algorithm, refer to Programming Challenges.Code:/*************************************************************************** *   Copyright (C) 2008 by Liu Kaipeng                                     * *   LiuKaipeng 

According to the Goldbach's Conjecture, every even number that is not less than 4 is the sum of two primes. So we can just reduce the number to an even number by subtracting 4=2+2 or 5=2+3 from it, and then test every possible summation of the

SIGIR-20102010.1.15 : Abstracts for full research papers due2010.1.22 : Full research paper submissions due2010.2.12 : Posters, demonstration, and tutorial proposals due2010.3.24 : All other acceptance notificationsKDD-20102010.2.2 : Abstract due 201