思路 分别统计这个位置左边和右边各有多少大于和小于它的数,乘起来即可 使用权值树状数组 代码 #include <cstdio> #include <algorithm> #include <cstring> #define int long long using namespace std; const int MAXN = 100000; struct BIT{ int a[MAXN+10]; int lowbit(int x){ return x&(-x)…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2492 Ping pong Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank…

                                  Ping pong Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description   N(3N20000) ping pong players live along a west-east street(consider the street as a li…

                                                                      Ping pong Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3139   Accepted: 1157 Description N(3<=N<=20000) ping pong players live along a west-east street(consider th…

N (3N20000)ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If two players want to compete, they must choose…

Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4094    Accepted Submission(s): 1522 Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the st…

题目链接:http://poj.org/problem?id=3928 乒乓比赛,有N个人参加,输入每个玩家的技能等级,对每个人设置一个特定ID和一个技能值,一场比赛需要两个选手和一个裁判,只有当裁判的ID和技能值都在两个选手之间的时候才能进行一场比赛,现在问一共能组织多少场比赛. 参考的其他人的代码,重新敲了一遍. /*POJ 3928 Ping pong */ #include<cstdio> #include<cstring> #include<algorithm>…

Frequent values 题意是不同颜色区间首尾相接,询问一个区间内同色区间的最长长度. 网上流行的做法,包括翻出来之前POJ的代码也是RMQ做法,对于序列上的每个数,记录该数向左和向右延续的最远位置,那么对于一个查询Q(L, R),它的答案就分成了三种情况right(L) - L,R - left(R)以及Q(L+right(L),R-left(R)). 这里给出一个线段树做法,在线段树的节点上维护3个量:l_value, r_value, value分别表示以左端点为起始点,以右端点为…

UVAlive 4329 Ping pong 题目: Ping pong Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description N(3N20000) ping pong players live along a west-east street(consider the street as a line segment). Each player ha…

Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4874    Accepted Submission(s): 1777 Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the st…

hdu2492 Ping pong 题意:一群乒乓爱好者居住在一条直线上,如果两个人想打比赛需要一个裁判,裁判的 位置 必须在两者之间 ,裁判的能力也必须不大于 参赛者最大的,不小于参赛者最小的 白皮的题解:考虑 第 i 个 为裁判 的情况 如果 左边 比 a[i] 小的 人数 为 c[i],则 有  i-c[i]-1 个人 比 a[i] 大,同样 右边 比 a[i] 小 的人数为 d[i] ,则比a[i]大的人为 n - i - d[i]; 则方案 为 d[i]*( i - c[i] - 1…

http://poj.org/problem?id=3928 Ping pong Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2087   Accepted: 798 Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player…

主题链接: PKU:http://poj.org/problem?id=3928 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2492 Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To im…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2492 Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4961    Accepted Submission(s): 1811 Problem Description N(3<=N<=20000) ping pong p…

Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2611    Accepted Submission(s): 973 Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the str…

近期在看Unity官方的Stealth项目教学视频,视频使用的是Unity旧的版本号,而我如今正在使用的是Unity 4.5,动画系统的操作全然不同了. 依照视频的方式根本无法设置动画的Ping Ping效果. 于是到官方的社区搜索了一番,发现有一篇帖子攻克了我的问题.如今把他的方法提供出来: 首先依照视频中解说的方式.创建cctvSweep的动画文件. 接下来.选择你想要做动画的对象(在视频中就是prop_cctvCam_joint对象),在Inspector视图点击Add Component…

Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If two players want to compete,…

Ping pong Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2302   Accepted: 879 Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To im…

欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4589    Accepted Submission(s): 1681 Problem Description N(3<=N<=20000) ping pong players live along a…

  Time Limit: 1000MS   Memory Limit: 65536KB   64bit IO Format: %lld & %llu Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their…

[Description] N (3 ≤ N ≤ 20000) ping pong players live along a west-east street(consider the street as a line segment).Each player has a unique skill rank. To improve their skill rank, they often compete with each other. Iftwo players want to compete…

erlang分布式入门(一)-ping pong 测试环境和http://willvvv.iteye.com/blog/1523918 一样,192.168.0.182(centos-182)和192.168.0.183(centos-183), 1.按照上面链接(步骤1-4)设置通过ssh hostname 免输入密码直接登录. 2.设置erlang的magiccookie,由于都是以root账号操作,分别在两台机子上执行以下命令 cd /root vi .erlang.cookie 输入 c…

UVA 1428 - Ping pong 题目链接 题意:给定一些人,从左到右,每一个人有一个技能值,如今要举办比赛,必须满足位置从左往右3个人.而且技能值从小到大或从大到小,问有几种举办形式 思路:利用树状数组处理出每一个位置左边比它小的个数和右边比他小的个数和.那么左边和右边大就也能计算出来,那么比赛场次为左边小*右边大+左边大*右边小. 代码: #include <cstdio> #include <cstring> const int N = 100005; int t,…

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2330 参考资料:<算法入门经典训练指南>刘汝佳 P197 这本书上面写的题目大意.解题思路都写出来了. 在这我贴上自己的 AC代码: #include<stdio.h> #include<string.h> #define PEOPLE 2…

对于我这样一名脑残ACMer选手,这道题看了好久好久大概4天,终于知道怎样把它和“树状数组”联系到一块了. 树状数组是什么意思呢?用十个字归纳它:心里有数组,手中有前缀. 为什么要用树状数组?假设你要储存一段数字的前缀和,还要动态修改这些数字.怎么办? 通常的想法就是用数组a[]保存所有的数字,再用数组s[]保存每一位上的前缀和. ;i<=n;i++){ s[i]=s[i-]+a[i]; } 这样有一个弊端,就是当你动态修改a[]数组中的数字的时候,维护s[]数组的代价太高了,最坏可达O(N),…

题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=13895 题意:一条街上住有n个乒乓选手,每个人都有一个技能值,现在每场比赛需要3个人,两个选手,一个裁判:裁判须住在他们之间,且其技能值必须在两选手之间,求一共能组织多少种比赛. 注意到技能值的范围(1 ≤ ai ≤ 100000),所以我们可以树状数组(O(nlogn))处理得到一个ll数组,ll[x]表示住在x左边的并且技能值小于x的技能值的人数.类似逆着处理…

题目传送门 题意:训练指南P197 分析:枚举裁判的位置,用树状数组来得知前面比它小的和大的以及后面比它小的和大的,然后O (n)累加小 * 大 + 大 * 小 就可以了 #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 1e5 + 5; const int M = 2e4 + 5; struct BIT { int c[N], sz; void init(int n) { mem…

题目大意:有n个数排成一列,问从中能找出几个三元组(ai,aj,ak)满足i<j<k并且这三个数严格单调. 题目分析:枚举中间的数字aj,如果aj前面有c(j)个数a(j)小,后面有d(j)个数比a(j)小,那么aj为中间数时,共有c(j)*(n-j-d(j))+d(j)*(j-1-c(j)).定义x(i)表示 i 出现过几次,当枚举到aj 时,只需统计sum(x(1)~x(a(j)-1)便得c(j),d(j)求法类似. 代码如下: # include<iostream> # in…

第一次写树状数组,感觉那个lowbit位运算用的相当厉害. 因为-x相当于把x的二进制位取反然后整体再加上1,所以最右边的一个1以及末尾的0,取反加一以后不变. 比如1000取反是0111加一得到1000,这样与运算以后不变 最右边的1左边部分取反,加一不会影响左半部分,所以与运算以后全部为0 对于这道题来说貌似不是很容易能联想到树状数组 注意题中说了每个人的技能值互不相同. 从左往右扫描每个a[i],另x[a[i]] = 1,然后统计x[1]...x[a[i]-1]的和就是第i个人左边技能值比…

题意:一条大街上住着n个乒乓球爱好者,经常组织比赛切磋技术.每个人都有一个能力值a[i].每场比赛需要三个人:两名选手,一名裁判.他们有个奇怪的约定,裁判必须住在两名选手之间,而裁判的能力值也必须在两名选手之间.问一共能组织多少种比赛. 分析:考虑第i个人,假设a1到ai-1中有ci个比ai小,那么就有(i-1)-ci个比ai大:同理,如果ai+1到an中有di个比ai小,那么就有(n-i)-di个比ai大.i当裁判就有:ci * (n-i-di) + (i-1-ci)*di种比赛. 然后问题就…