hdu 3874 Necklace(线段树)】的更多相关文章

欢迎參加--每周六晚的BestCoder(有米! ) Necklace Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4003    Accepted Submission(s): 1330 Problem Description Mery has a beautiful necklace. The necklace is made…
Necklace HDU - 3874  Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count…
Necklace Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2083    Accepted Submission(s): 747 Problem Description Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ba…
Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define t…
在所有数字的统计范围,,对于重复统计只有一次 离线段树算法 排序终点坐标.然后再扫,反复交锋.把之前插入树行被删除 #include "stdio.h" #include "string.h" #include "algorithm" using namespace std; struct node { int l,r; __int64 sum; }data[200010]; struct Mark { int l,r,id; }mark[200…
Attack Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 2496    Accepted Submission(s): 788 Problem Description Today is the 10th Annual of “September 11 attacks”, the Al Qaeda is about to attack…
Coder Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4838    Accepted Submission(s): 1853 Problem Description In mathematics and computer science, an algorithm describes a set of procedures…
Man Down Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2030    Accepted Submission(s): 743 Problem Description The Game “Man Down 100 floors” is an famous and interesting game.You can enjoy t…
1.HDU 5877  Weak Pair 2.总结:有多种做法,这里写了dfs+线段树(或+树状树组),还可用主席树或平衡树,但还不会这两个 3.思路:利用dfs遍历子节点,同时对于每个子节点au,查询它有多少个祖先av满足av<=k/au. (1)dfs+线段树 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm>…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3308 题目很好懂,就是单点更新,然后求区间的最长上升子序列. 线段树区间合并问题,注意合并的条件是a[mid + 1] > a[mid],写的细心点就好了. #include <iostream> #include <cstring> #include <cstdio> using namespace std; ; struct SegTree { int l , r…