题意: 给出一个序列和若干次询问,每次询问一个子序列去重后的所有元素之和. 分析: 先将序列离散化,然后离线处理所有询问. 用莫队算法维护每个数出现的次数,就可以一边移动区间一边维护不同元素之和. #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long LL; const int maxn…

http://acm.hdu.edu.cn/showproblem.php?pid=3333 Turing Tree Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2614    Accepted Submission(s): 892 Problem Description After inventing Turing Tree, 3…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6278 Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others) Problem Description The h-index of an author is the largest h where he has at least h papers with citations not les…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3333 Turing Tree Time Limit: 6000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) 问题描述 After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because…

Turing Tree Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5014    Accepted Submission(s): 1777 Problem Description After inventing Turing Tree, 3xian always felt boring when solving problems…

就是莫队的模板题 /* Memory: 0 KB Time: 1663 MS Language: C++11 4.8.2 Result: Accepted */ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cstdlib> #include<cmath> using namespace std; typedef l…

题目链接 给n个数, m个询问, 每次询问输出区间内的数的和, 相同的数只计算一次. 数组里的数是>-1e9 <1e9, 可以把它离散以后用莫队搞... #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #i…

Turing Tree Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4768    Accepted Submission(s): 1686 Problem Description After inventing Turing Tree, 3xian always felt boring when solving problems…

题意:给一个数组,每次查询输出区间内不重复数字的和. 这是3xian教主的题. 用前缀和的思想可以轻易求得区间的和,但是对于重复数字这点很难处理.在线很难下手,考虑离线处理. 将所有查询区间从右端点由小到大排序,遍历数组中的每个数字,每次将该数字上次出现位置的值在树状数组中改为0,再记录当前位置,在树状数组中修改为当前的数值.这样可以保证在接下来的查询中该数字只出现了一次.这是贪心的思想,只保留最可能被以后区间查询的位置.如果当前位置是某个查询区间的右端点,这时候就可以查询了.最后再根据查询区间…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3333 题意就是询问区间不同数字的和. 比较经典的树状数组应用. //#pragma comment(linker, "/STACK:102400000, 102400000") #include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #inc…