其他

Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V’, E’), with

Description In mathematics, a squarefree number is one which is divisible by no perfect squares, except 1. For example, 10 is square-free but 18 is not, as it is divisible by 9 = 3^2. Now you need to determine whether an integer is squarefree

Frogs' Neighborhood Time Limit: 5000MS   Memory Limit: 10000K Total Submissions: 8348   Accepted: 3538   Special Judge

描述 對正整數 n ，歐拉函數是小於或等於 n 的數中與 n 互質的數的數目。此函數以其首名研究者歐拉命名，它又稱為 Euler’s totient function 、 φ 函數、歐拉商數等。例如：φ(8) = 4（Phi(8) = 4），因為 1,3,5,7 均和 8 互質。 S(n) = Phi(1) + Phi(2) + …… Phi(n)，給出n，求S(n)，例如：n = 5，S(n) = 1 + 1 + 2 + 2 + 4 = 10，定義Phi(1) =

The Ghost Blows Light Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2549    Accepted Submission(s): 795 Problem Description My

Description King OMeGa catched three men who had been streaking in the street. Looking as idiots though, the three men insisted that it was a kind of performance art, and begged the king to free them. Out of hatred to the real idiots, the king

類和結構可包含比整數型別佔用更少儲存空間的成員。這些成員被指定為位域。位域成員聲明符規範的文法如下： 文法 declarator : constant-expression

2016年7月28日星期四 T.s.road總結筆記：分組背包問題Matlab實現——之基本背包擴充  項目源碼：https://github.com/Tsroad/KnapsackProblemSeries 作者說明： Whenrunning thisprogramme, the author’s PC setting is: MicrosoftWindows 7 (SP1) + Matlab R2010b +CPU i5-4590 + RAM

Description In mathematics, the function d(n) denotes the number of divisors of positive integer n. For example, d(12)=6 d(12)=6 because 1,2,3,4,6,12 1,2,3,4,6,12 are all 12 12 ‘s

Rescue The Princess Time Limit: 1000ms   Memory limit: 65536K  有疑問。點這裡^_^ 題目描述     Several days ago, a beast caught a beautiful princess and the princess was put in prison. To rescue

題意：就是給出n和b，然後給出n個數，用這n個數中的某些，求出一個和，這個和是>=b的最小值，輸出最小值與b的差。 分析：這道題很簡單，是很明顯的01背包問題，這裡的n個物品，每個物品的重量為c[i]，價值為w[i]並且c[i]==w[i]，，容量為所有c[i]的和sum。只要在f[]中從頭開始找，找到一個最小>=b的就是題目要的解   Description Farmer John recently bought

Description Suppose some supervisors each get to hire a new person for their department. There are N people to be placed in these N departments. Each supervisor interviews all N people, and ranks them according to how much she wants each of

http://codeforces.com/contest/887/problem/D 一個模特有兩種活動。 ① 拍照片，掙錢 a。 ②開演唱會，花費b 給定模特這兩種工作的時間表。 模特可以選定一個時間舉辦一個座談會，那麼他拍照片的錢變c。開演唱會會花費d。 要求再模特座談會之前和後len天 都不能賠錢。 要求你輸出最小的座談會天數。 沒有輸出-1.。 思路：枚舉第i天作為 開始的茶話會的時間（茶話會是可以和當天其他的活動撞的，並且當天有影響），然後計算後len天是否有負數。

Problem Description In a highly developed alien society, the habitats are almost infinite dimensional space. In the history of this planet,there is an old puzzle. You have a line segment with x units’ length representing one

Interceptors 　　攔截器(Interceptors)是一種強有力的途徑，來監控，改寫和重試HTTP訪問。下面是一個簡單的攔截器，對流出的請求和流入的響應記錄日誌。 class LoggingInterceptor implements Interceptor { @Override public Response intercept(Chain chain) throws IOException { Request request = chain.request(

描述 莫比烏斯函數，由德國數學家和天文學家莫比烏斯提出。梅滕斯(Mertens)首先使用μ(n)（miu(n)）作為莫比烏斯函數的記號。具體定義如下： 如果一個數包含平方因子，那麼miu(n) = 0。例如：miu(4), miu(12), miu(18) = 0。 如果一個數不包含平方因子，並且有k個不同的質因子，那麼miu(n) = (-1)^k。例如：miu(2), miu(3), miu(30) = -1,miu(1), miu(6), miu(10) = 1。

Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following: Choose k different positive integers a1, a2, …, ak. For some non-negative m,

1 資料存放區   1.1 增加一條採樣資料 gnocchi --debug measures add -m 2016-04-16T14:33:58@43.1 --resource-id 3068f5ec-0d37-419a-929d-0505efaf908c vm_meteric_001   1.2 Rest api處理介面

Description Function Fx,y F_{x,y} satisfies: F1,1=F1,2=1 F_{1,1}=F_{1,2}=1 F1,i=F1,i−1+2∗F1,i−2 (i>=3) F_{1,i}=F_{1,i-1}+2*F_{1,i-2}\ (i>=3) Fi,j=∑j+N−1k=jFi−1,

我又來水部落格了。 分組背包是LC(01)背包的一種變形。分組背包中物品被分成幾組，每組中只能挑選出一件物品加入背包，這是與01背包的區別。 在01背包中，我們以每一件物品作為動態規劃的每一階段，但是在分組背包中我們要以每一組作為每一階段。 其實很簡單，代碼如下。 for(int i=1;i<=k;i++)for(int c=v;c>=0;c--)for( each 物品j in 第i組 )if(c>=w[i])f[c]=max(f[c],f[c-w[j]+v[j]]