1. Oracle: "MERGE into MHGROUP.proj_access m using dual on " + "(PRJ_ID = '" + WS_PrjID + "' AND USER_GP_ID = '" + smrIDs[i] + "')" + "when not matched then " + "insert (PRJ_ID,OBJECT_TYPE,USER_GP_ID,…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5458 Problem Description Given an undirected connected graph G with n nodes and m edges, with possibly repeated edges and/or loops. The stability of connectedness between node u and node v is defined by…
2962: 序列操作 Time Limit: 50 Sec Memory Limit: 256 MBSubmit: 678 Solved: 246[Submit][Status][Discuss] Description 有一个长度为n的序列,有三个操作1.I a b c表示将[a,b]这一段区间的元素集体增加c,2.R a b表示将[a,b]区间内所有元素变成相反数,3.Q a b c表示询问[a,b]这一段区间中选择c个数相乘的所有方案的和mod 19940417的值. Input 第一…
3638: Cf172 k-Maximum Subsequence Sum Time Limit: 50 Sec Memory Limit: 256 MBSubmit: 174 Solved: 92[Submit][Status][Discuss] Description 给一列数,要求支持操作: 1.修改某个数的值 2.读入l,r,k,询问在[l,r]内选不相交的不超过k个子段,最大的和是多少. Input The first line contains integer n (1 ≤ n …
E. Maze 2D time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The last product of the R2 company in the 2D games' field is a new revolutionary algorithm of searching for the shortest path in…
HDU 3397 Sequence operation 题目链接 题意:给定一个01序列,有5种操作 0 a b [a.b]区间置为0 1 a b [a,b]区间置为1 2 a b [a,b]区间0变成1,1变成0 3 a b 查询[a,b]区间1的个数 4 a b 查询[a,b]区间连续1最长的长度 思路:线段树线段合并.须要两个延迟标记一个置为01,一个翻转,然后因为4操作,须要记录左边最长0.1.右边最长0.1,区间最长0.1,然后区间合并去搞就可以 代码: #include <cstdi…