1100. Final Standings Time limit: 1.0 secondMemory limit: 16 MB Old contest software uses bubble sort for generating final standings. But now, there are too many teams and that software works too slow. You are asked to write a program, which generate…

http://acm.timus.ru/problem.aspx?space=1&num=1100 link to the problem make a fast stable sorting algorithm. what is sort in c, quick sort. what is a stable sort? a sorting algorithm is said to be stable if two objects with equal keys appear in the sa…

一.final final的中文意思就是不可更改的,最终的. 1.final修饰变量,那么该变量无法更改.一旦该变量赋了初值,就不能重新赋值. final MAX = 1100; //final修饰后MAX就无法重新赋值. 2.final修饰方法 final修饰方法,那么该方法不能被子类重写,但可以重载. class Person{ final void getName(){.....} } //getName()方法,无法被子类重写. 3.final修饰类 final修饰类,该类无法被继承.…

修改 Makefile DCNN=1 -> DCNN=0 // 禁用DCNN #BOARD_SIZE=19 -> BOARD_SIZE=19 // 棋盘大小19x19 OPT ?= -O3 -> OPT ?= -O0 // 优化gcc编译选项 运行参数: prior=eqex=0,dynkomi=none 参数pachi.c int debug_level = 3; -> int debug_level = 5; // 内部debug等级,主要用来打印相关信息uct.c u->…

Problem C. Contest 题目连接: http://codeforces.com/gym/100714 Description The second round of the annual student collegiate programming contest is being held in city N. To be prepared for the inrush of participants, the jury needs to know the number of t…

PAT 1139 1138 1137 1136 一个月不写题,有点生疏..脑子跟不上手速,还可以啦,反正今天很开心. PAT 1139 First Contact 18/30 找个时间再修bug 23/30 28/30 30/30 补完啦 这道题的细节坑点: 1.输出id需要补全4位 用print("%04d") 5分ok 2.需要舍弃查询的相恋男女的直接边 3分ok 3.大坑 0000 与 -0000 使用int型无法区分男女的. 2分ok 想法:用string输入 看长度和第一位符…

链接: https://codeforces.com/contest/1265/problem/C 题意: So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved pi problems. Since…

1 Problem Description Problem The final match of the Berland Football Cup has been held recently. The referee has shown n yellow cards throughout the match. At the beginning of the match there were a1 players in the first team and a2 players in the s…

Description \(n,q,V\leq 100000,w_i\leq 10^9\) Solution 又是一道大数据结构 由于有一个下取整,这就导致了不同时间的修改值是不能简单的直接加在一起的. 容易发现,1操作的影响只会影响到距离不超过log的点. 这样我们很容易得到一个\(q\log n\log ^2V\)的做法 同一深度的修改有一种套路是维护BFS序. 对于子树内的点,我们将log个深度对应的BFS序区间减去相应的影响. 对于修改点的log个有用的祖先,我们也类似操作,注意重复影响…

接着上一篇Blog:一道面试题与Java位操作 和 BitSet 库的使用,分析下Java源码中BitSet类的源码. 位图(Bitmap),即位(Bit)的集合,是一种常用的数据结构,可用于记录大量的0-1状态,在很多地方都会用到,比如Linux内核(如inode,磁盘块).Bloom Filter算法等,其优势是可以在一个非常高的空间利用率下保存大量0-1状态.在Java中,直接面向程序员的最小数据操作粒度是byte,并不提供能够直接操作某个bit的途径,但是程序员可以通过使用位运算符(&…