摘要： SeasideTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 396Accepted Submission(s): 272Problem DescriptionXiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives i 阅读全文

摘要： Permutation CountingTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 493Accepted Submission(s): 258Problem DescriptionGiven a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the 阅读全文

摘要： AssignmentsTime Limit: 4000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 645Accepted Submission(s): 300Problem DescriptionIn a factory, there are N workers to finish two types of tasks (A and B). Each type has N tasks. Each task of type A needs xi time to finish 阅读全文

摘要： Least Common MultipleTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11716Accepted Submission(s): 4278Problem DescriptionThe least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the nu 阅读全文

摘要： Uniform GeneratorTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6215Accepted Submission(s): 2477Problem DescriptionComputer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the formseed(x+1) 阅读全文

摘要： Digital RootsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20786Accepted Submission(s): 6161Problem DescriptionThe digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then tha 阅读全文

摘要： u Calculate eTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12753Accepted Submission(s): 5485Problem DescriptionA simple mathematical formula for e iswhere n is allowed to go to infinity. This can actually yield very accurate approximations of e u 阅读全文

摘要： ElevatorTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16549Accepted Submission(s): 8795Problem DescriptionThe highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floor 阅读全文

摘要： Tempter of the BoneTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23769Accepted Submission(s): 6538Problem DescriptionThe doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, a 阅读全文

摘要： 最近又复习了下最大流问题，每次看这部分的内容都会有新的收获。可以说最大流问题的资料网上一搜一大把，根本没有必要自己写；但是大部分资料上的专业术语太多了，初学很难理解，至少我当年学这部分的时候前几次就没有看懂。所以我准备备份一点个人的理解。图-1 如图-1所示，在这个运输网络中，源点S和汇点T分别是1,7，各边的容量为C(u,v)。图中红色虚线所示就是一个可行流。标准图示法如图-2所示：其中p(u,v) / c(u,v)分别表示该边的实际流量与最大容量。 关于最大流 熟悉了什么是网络流，最大流也就很好理解了。就是对于任意的u∈V-{s},使得p(s,u)的和达到最大。上面的运输网络中，最大流如图 阅读全文