其他

So Easy!Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 899 Accepted Submission(s): 244Problem Description　　A sequence Sn is defined as:Where a, b, n, m are positive integers.┌x┐is the ceil of x.

The Little Girl who Picks MushroomsTime Limit: 2 Seconds Memory Limit: 32768 KB It's yet another festival season in Gensokyo. Little girl Alice planned to pick mushrooms in five mountains. She brought five bags with her and used different bags to

Maximum Random WalkTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 229 Accepted Submission(s): 128Problem DescriptionConsider the classic random walk: at each step, you have a 1/2 chance of taking

lockerTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 880 Accepted Submission(s): 371Problem DescriptionA password locker with N digits, each digit can be rotated to 0-9 circularly.You can rotate 1

Aeroplane chessTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 833 Accepted Submission(s): 575Problem DescriptionHzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0

Travel in timeTime Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1008 Accepted Submission(s): 204Problem Description　　Bob gets tired of playing games, leaves Alice, and travels to Changsha alone.

Tree CuttingTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3310 Accepted: 1952DescriptionAfter Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible cost, he

Breaking StringsTime Limit: 2 Seconds      Memory Limit: 65536 KBA certain string-processing language allows the programmer to break a string into two pieces. Since this involves copying the old string, it costs n units of time to break a string of

所謂最小二乘法，就是對一系列的二維觀測值進行直線擬合，假設該直線為Y = kx + b, 那麼如何評判這個條直線是否能很好地擬合所有的觀測值呢？最小二乘法的方法就是如果k，b使得 通過該直線計算出來的值Y和觀測值yi的差的平方和最小，那麼就可以認為該直線可以擬合觀測值 即使得下式值最小我們可以分別對k和b進行求偏導，使得兩個偏導均為0即可，即如下兩式上面兩個等式其實就相當於是一個二元一次方程，通過求解，可以得到k，b如下：           

Balancing ActTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7594 Accepted: 3082DescriptionConsider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more

Scout YYF ITime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4023 Accepted: 1029DescriptionYYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is

#!/usr/bin/pythonfrom math import sqrtdef genUserBasedMap(file = 'u.data'): map = {} f = open(file) for line in f: (user, item, rate) = line.split('\t')[0:3] map.setdefault(int(user), {}) map[int(user)][int(item)] = int(

Collecting BugsTime Limit: 10000MS Memory Limit: 64000KTotal Submissions: 1730 Accepted: 794Case Time Limit: 2000MS Special JudgeDescriptionIvan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he

Sum of divisorsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2052 Accepted Submission(s): 713Problem Descriptionmmm is learning division, she's so proud of herself that she can figure out the

One Person GameTime Limit: 1 Second Memory Limit: 32768 KB Special Judge There is a very simple and interesting one-person game. You have 3 dice, namelyDie1, Die2 and Die3. Die1 hasK1 faces. Die2 has K2 faces.Die3 has K3 faces. All the dice are fair

charge-stationTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 610 Accepted Submission(s): 306Problem DescriptionThere are n cities in M^3's empire. M^3 owns a palace and a car and the palace

RSA FactorizationTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 471 Accepted: 124DescriptionThe positive integer n is given. It is known that n = p * q, where p and q are primes, q <= p and |q - kp| <= 105 for some given positive

“一分鐘速算口訣”：兩位元相乘，在十位元相同、個位元相加等於10的情況下，如62×68=4216計算方法：6×（6+1）=42（前積），2×8=16（後積）。一分鐘速算口訣中對特殊題的定理是：任意兩位元乘以任意兩位元，只要魏式係數為“0”所得的積，一定是兩項數中的尾乘尾所得的積為後積，頭乘頭（其中一項頭加1的和）的積為前積，兩積相鄰所得的積。如（1）33×46=1518（個位元相加小於10，所以十位元小的數字3不變,十位大的數4必須加1）計算方法：3×（4+1）=15（前積），3×6=18（後積

矩陣 行列式數對順序逆序逆序數偶/奇排列上三角行列式 = 下三角行列式 =  對稱三角行列式 = 主對角線的乘積轉置：行列式中的行和列互換交換行列式中兩行的所有元素，即原行列式的值為D1，則新的值為-D1     =》 如果行列式中有兩行的完全一樣，那麼行列式為0行列式的某行加上另一行對應元素的K倍，行列式的值不變餘子式  代數n階行列式 = 任一行的個元素與它們對應的代數餘子式的乘積之和行列式的計算：        使得某一行或者某一列中有盡量多的0，然後應用上面的公式       

Yukari's BirthdayTime Limit: 2 Seconds Memory Limit: 32768 KB Today is Yukari's n-th birthday. Ran and Chen hold a celebration party for her. Now comes the most important part, birthday cake! But it's a big challenge for them to placen candles on