Tug of War(A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other the number of people on the two teams must not differ by more than 1 the total weight of the people on each team should be as nearly equal as possible. The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1 the second the weight of person 2 and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic. Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first. )
上传时间： 2014-01-06
上传用户：离殇
This is a package to calculate Discrete Fourier/Cosine/Sine Transforms of 1-dimensional sequences of length 2^N. This package contains C and Fortran FFT codes.
标签： dimensional Transforms calculate sequences
上传时间： 2014-01-14
上传用户：LIKE
一个基于NHibernate的N层开发基础框架（可以，马上应用到你的项目中），使用.NET（C＃）开发。基本操作CRUD完全实现，数据间的关系(one-to-many,many-to-many)均有实现!
标签： NHibernate 开发基础
上传时间： 2013-12-19
上传用户：gaojiao1999
．数据结构 假设有M个进程N类资源，则有如下数据结构： MAX[M*N] M个进程对N类资源的最大需求量 AVAILABLE[N] 系统可用资源数 ALLOCATION[M*N] M个进程已经得到N类资源的资源量 NEED[M*N] M个进程还需要N类资源的资源量 2．银行家算法 设进程I提出请求Request[N]，则银行家算法按如下规则进行判断。 (1)如果Request[N]<=NEED[I，N]，则转(2)；否则，出错。 (2)如果Request[N]<=AVAILABLE，则转(3)；否则，出错。 (3)系统试探分配资源，修改相关数据： AVAILABLE=AVAILABLE-REQUEST ALLOCATION=ALLOCATION+REQUEST NEED=NEED-REQUEST (4)系统执行安全性检查，如安全，则分配成立；否则试探险性分配作废，系统恢复原状，进程等待。 3．安全性检查 (1)设置两个工作向量WORK=AVAILABLE；FINISH[M]=FALSE (2)从进程集合中找到一个满足下述条件的进程， FINISH[i]=FALSE NEED<=WORK 如找到，执行(3)；否则，执行(4) (3)设进程获得资源，可顺利执行，直至完成，从而释放资源。 WORK=WORK+ALLOCATION FINISH=TRUE GO TO 2 (4)如所有的进程Finish[M]=true，则表示安全；否则系统不安全。
上传时间： 2014-01-05
上传用户：moshushi0009
数据结构 假设有M个进程N类资源，则有如下数据结构： MAX[M*N] M个进程对N类资源的最大需求量 AVAILABLE[N] 系统可用资源数 ALLOCATION[M*N] M个进程已经得到N类资源的资源量 NEED[M*N] M个进程还需要N类资源的资源量 2．银行家算法 设进程I提出请求Request[N]，则银行家算法按如下规则进行判断。 (1)如果Request[N]<=NEED[I，N]，则转(2)；否则，出错。 (2)如果Request[N]<=AVAILABLE，则转(3)；否则，出错。 (3)系统试探分配资源，修改相关数据： AVAILABLE=AVAILABLE-REQUEST ALLOCATION=ALLOCATION+REQUEST NEED=NEED-REQUEST (4)系统执行安全性检查，如安全，则分配成立；否则试探险性分配作废，系统恢复原状，进程等待。 3．安全性检查 (1)设置两个工作向量WORK=AVAILABLE；FINISH[M]=FALSE (2)从进程集合中找到一个满足下述条件的进程， FINISH[i]=FALSE NEED<=WORK 如找到，执行(3)；否则，执行(4) (3)设进程获得资源，可顺利执行，直至完成，从而释放资源。 WORK=WORK+ALLOCATION FINISH=TRUE GO TO 2 (4)如所有的进程Finish[M]=true，则表示安全；否则系统不安全。
上传时间： 2013-12-23
上传用户：alan-ee
ENGLISH RESUME Dear sir/madam In answer to your advertisement in present interview for your need. I wish to tender my service. With reference to your advertisement in present interview for your need, I respectfully offer myself for the position.
标签： your advertisement interview ENGLISH
上传时间： 2014-09-08
上传用户：chenbhdt
Rotating shafts experience a an elliptical motion called whirl. It is important to decompose this motion into a forward and backward whil orbits. The current function makes use of two sensors to generate a bi-directional spectrogram. The method can be extended to any time-frequency distribution ％ ％ compute the forward/backward Campbell/specgtrogram ％ ％ INPUT: ％ y (n x 2) each column is measured from a different sensor ％ /////// ％ __ ％ |s1| y(:,1) ％ |__| ％ __ ％ / \ ________|/ ％ | | | s2 |/ y(:,2) ％ \____/ --------|/ ％ ％ Fs Sampling frequnecy ％ ％ OUTPUT: ％ B spectrogram/Campbel diagram ％ x x-axis coordinate vector (time or Speed) ％ y y-axis coordinate vector (frequency [Hz])
标签： experience elliptical decompose important
上传时间： 2015-06-23
上传用户：372825274
contain many examples code for I2c,UART,string ,digital convert, read/write to EEprom in microchip PIC16xx series. and the default compiler is hitech PIC16.
标签： microchip examples contain digital
上传时间： 2014-01-04
上传用户：xmsmh
/* * EULER S ALGORITHM 5.1 * * TO APPROXIMATE THE SOLUTION OF THE INITIAL VALUE PROBLEM: * Y = F(T,Y), A<=T<=B, Y(A) = ALPHA, * AT N+1 EQUALLY SPACED POINTS IN THE INTERVAL [A,B]. * * INPUT: ENDPOINTS A,B INITIAL CONDITION ALPHA INTEGER N. * * OUTPUT: APPROXIMATION W TO Y AT THE (N+1) VALUES OF T. */
标签： APPROXIMATE ALGORITHM THE SOLUTION
上传时间： 2015-08-20
上传用户：zhangliming420
American Gladiator,You are consulting for a game show in which n contestants are pitted against n gladiators in order to see which contestants are the best. The game show aims to rank the contestants in order of strength this is done via a series of 1-on-1 matches between contestants and gladiators. If the contestant is stronger than the gladiator, then the contestant wins the match otherwise, the gladiator wins the match. If the contestant and gladiator have equal strength, then they are “perfect equals” and a tie is declared. We assume that each contestant is the perfect equal of exactly one gladiator, and each gladiator is the perfect equal of exactly one contestant. However, as the gladiators sometimes change from one show to another, we do not know the ordering of strength among the gladiators.
标签： contestants consulting Gladiator are
上传时间： 2013-12-18
上传用户：windwolf2000
