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,…

树状数组,把他们的技能值作为轴: 首先按照编号从小到大插入值,这样就可以得到,技能值比当前小的人数: 然后按照编号从大到小再插一遍: 代码: #include<cstdio> #include<cstring> #define maxn 20005 using namespace std; ]; int l[maxn],r[maxn]; int lowbit(int x){return x&-x;} ){c[x]+=d,x+=lowbit(x);}} int sum(int…

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…