solved 7/11 2016 Multi-University Training Contest 10 题解链接 分类讨论 1001 Median(BH) 题意: 有长度为n排好序的序列,给两段子序列[l1,r1],[l2,r2]构成新的序列,问中间的数字. 思路: 根据不同情况分类讨论即可.时间复杂度O(1). 代码: #include <bits/stdc++.h> const int N = 1e5 + 5; int a[N]; int n, m; int l1, r1, l2, r…

CRB and Tree                                                             Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                                            To…

CRB and Apple Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 421    Accepted Submission(s): 131 Problem Description In Codeland there are many apple trees.One day CRB and his girlfriend decide…

CRB and Queries Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 533    Accepted Submission(s): 125 Problem DescriptionThere are N boys in CodeLand.Boy i has his coding skill Ai.CRB wants to k…

Welcome Party Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 875    Accepted Submission(s): 194 Problem Description The annual welcome party of the Department of Computer Science and Technolo…

2015 Multi-University Training Contest 10 5406 CRB and Apple 1.排序之后费用流 spfa用stack才能过 //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<bits/stdc++.h> using namespace std; function<…

HDU-4698 Counting 题意:给定一个二维平面,其中x取值为1-N,y取值为1-M,现给定K个点,问至少包括K个点中的一个的满足要求的<Xmin, Xmax, Ymin, Ymax>共有多少中取值情况.也就是说K个点中至少一个点落在所给定的区间内. 解法:正面求解,由于点只有1000个,因此直接暴力离散化之后的x轴坐标,对于y轴则可以通过增加一个一个加入点,使用一个set来维护纵轴有多少种不同的取法. #include <cstdlib> #include <cs…

Y Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 304    Accepted Submission(s): 104 Problem Description   Sample Input 4 1 2 1 3 1 4   Sample Output 1 Hint 1. The only set is {2,3,4}. 2. Ple…

Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 127    Accepted Submission(s): 60 Problem Description   Sample Input 2   Sample Output 2 解题思路: 很容易看得出,这是个组合数学的插板问题,答案为2^(n-1); 由于n特别大,则(2^(…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=5860 题目大意:给你n个人排成一列编号,每次杀第一个人第i×k+1个人一直杀到没的杀.然后剩下的人重新编号从1-剩余的人数.按照上面的方式杀.问第几次杀的是谁. 分析 一轮过后和原来问题比只是人的编号发生变化,故可以转化为子问题求解,不妨设这n个人的编号是0~n-1,对于第i个人,如果i%k=0,那么这个人一定是第一轮出列的第i/k+1个人:如果i%k!=0,那么这个人下一轮的编号就是i…