分析:找一个区间里小于等于h的数量,然后这个题先离散化一下,很简单 然后我写这个题主要是熟悉一下主席树,其实这个题完全可以离线做,很简单 但是学了主席树以后,我发现,在线做,一样简单,而且不需要思考 (主席树大法好)无限仰慕 #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; ; const int INF=0…

#include<iostream> #include<string.h> #include<algorithm> #include<stdio.h> #include<vector> #define LL long long #define rep(i,j,k) for(int i=j;i<=k;i++) #define per(i,j,k) for(int i=j;i>=k;i--) #define pb push_back #d…

Super Mario Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1437    Accepted Submission(s): 690 Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability…

Super Mario Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3625    Accepted Submission(s): 1660 Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability…

题意:给一个数组,每次询问输出在区间[L,R]之间小于H的数字的个数. 此题可以使用划分树在线解决. 划分树可以快速查询区间第K小个数字.逆向思考,判断小于H的最大的一个数字是区间第几小数,即是答案.这一步可以使用二分搜索上界.时间复杂度是O(logn*logn). #include <iostream> #include <cstdio> #include <cstring> #include <stack> #include <algorithm&…

把数值和查询放在一起从小到大排序,纪录每个数值的位置,当遇到数值时就更新到树状数组中,遇到查询就直接查询该区间和. #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; ; struct node { int id; int L, R; int val; } qq[MAXN]; int N, Q; int cntQ;…

题目链接 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define N 100100 struct node { int l,r; }tree[*N]; int sorted[N]; ][N]; ][N]; void build(in…

Super Mario Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5101    Accepted Submission(s): 2339 Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping abilit…

Super Mario Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4090    Accepted Submission(s): 1883 Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=4417 题意: 给你段长为n的序列,有q个询问,每次询问区间[l.r]内有多少个数小于等于k 思路: 之前用分块写过类似的,不过为了练习下主席树,这里用主席树写了下.思路很简单 离线离散化处理下,每次插入一个数num时,在主席树上下标num+1,这样每次询问[l,r]中有多少个小于k的数的时候,我们只要找下标[1,k]的区间第R次修改后的总和减去第L-1次修改后的总值就可以得到了 实现代码: #inclu…