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…

Super Mario Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trou…

Super Mario Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5370    Accepted Submission(s): 2461 Problem Description Mario is world-famous plumber. His "burly" figure and amazing jumping a…

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): 6077    Accepted Submission(s): 2645 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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4417 题目大意:给你n个数,下标为0到n-1,m个查询,问查询区间[l,r]之间小于等于x的数有多少个. 写的时候逗比了...还是写的太少了.. 我们按照x从小到大排序来查询,然后找区间上的点,如果小于等于它就插入,然后看这个区间内插入了多少个点. 点也是可以排序的.. 详见代码: #include <cstdio> #include <cmath> #include <algo…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4417 题意很简单,给定一个序列求一个区间 [L, R,]中小于等于H的元素的个数. 好像函数式线段树可解吧,可弱弱的沙茶一直没弄懂其精髓,只好用树套树暴力碾压了 额树套树,线段树的每一个节点套一个sb树. 当查询[l,r]区间中的值小于等于H的个数,先用线段树找到相应的区间, 然后再查询该区间下对应的平衡树中小于等于H的个数,累加即可. 一直以为会超时,结果400+ms就过了,数据应该很弱吧(自己对…

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…