Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds and n trees. The i-th bird stands on the top of the i-th tree. The trees stand in straight line…

Gunner II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1740    Accepted Submission(s): 635 Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting ve…

题目传送门 /* 题意:查询x的id,每次前排的树倒下 使用lower_bound ()查找高度,f[i]记录第一棵高度为x树的位置,查询后+1(因为有序) */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; struct A { int v, id; bool operator < (cons…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=5233 简单题,stl水之... #include<algorithm> #include<iostream> #include<cstdlib> #include<cstdio> #include<set> using std::multiset; struct Node { int val, pos; Node(, ) :val(_val), po…

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5233 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=585&pid=1002 题解: 离散化之后,存在一张表里面(相同的值的id号用链表串起来,倒着存),每次查询完就把表首的删了,继续查. 之前离散的时候没有把查询的一起加进去,就一直t,估计是查询的时候很多是没有结果的,就会比较耗时间.…

 Gunner II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1724    Accepted Submission(s): 631 Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting…

Gunner II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1244    Accepted Submission(s): 486 Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting ve…

题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2],…

题目链接:Populating Next Right Pointers in Each Node II | LeetCode OJ Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only u…

原文标题:Functional Android (II): Collection operations in Kotlin 原文链接:http://antonioleiva.com/collection-operations-kotlin/ 原文作者:Antonio Leiva(http://antonioleiva.com/about/) 原文发布:2015-09-29 在简化代码方面,Lambda表达式是一个杰出的工具,而且还可以完成之前不可能完成的事.我们在这个系列文章的第一篇(Unlea…