UVA 12436 - Rip Van Winkle's Code option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=3867&mosmsg=Submission+received+with+ID+14331401" target="_blank" style="">题目链接 题意:区间改动一个加入等差数列,一个把区间设为某个…

Rip Van Winkle was fed up with everything except programming. One day he found a problem whichrequired to perform three types of update operations (A, B, C), and one query operation S over an arraydata[]. Initially all elements of data are equal to 0…

Rip Van Winkle was fed up with everything except programming. One day he found a problem whichrequired to perform three types of update operations (A, B, C), and one query operation S over an arraydata[]. Initially all elements of data are equal to 0…

题目大意:一个数组,四种操作: long long data[250001]; void A( int st, int nd ) { for( int i = st; i <= nd; i++ ) data[i] = data[i] + (i - st + 1); } void B( int st, int nd ) { for( int i = st; i <= nd; i++ ) data[i] = data[i] + (nd - i + 1); } void C( int st, int…

题意: long long data[250001]; void A( int st, int nd ) { for( int i = st; i \le nd; i++ ) data[i] = data[i] + (i - st + 1); } void B( int st, int nd ) { for( int i = st; i \le nd; i++ ) data[i] = data[i] + (nd - i + 1); } void C( int st, int nd, int x…

偶数时,中位数之间的数都是能够的(包含中位数) 奇数时,一定是中位数 推导请找初中老师 #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; int box[1000000]; int main() { //freopen("in","r"…

常见条码类型,如下: 1.Code 39 Code 39,又称为"Code 3 of 9",是非零售市场中最常用的格式,用于盘存和跟踪.Code 39码编码规则简单,误码率低.所能表示字符多等特点.此条码广泛应用于制造业.军事和医疗保健行业中.这种格式离散而且长度可变,接受以下 44 个字符: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ-.$/+%及“空格”星号 () 仅作起始/结束字符,不能用于条码正文. 2.Code 128 Code 128 是一种长…

Mysql 查询运行过程 大致分为4个阶段吧: 语法分析(sql_parse.cc<词法分析, 语法分析, 语义检查 >) >>sql_resolver.cc # JOIN.prepare 生成逻辑查询plan(sql_optimizer.cc) >># JOIN.optimize 生成物理查询plan(sql_planner.cc) run the explain plan(sql_executor.cc) JOIN.exec JOIN.prepare() : 子查询…

话说STL的I/O流用的还真不多,就着这道题熟练一下. 用了两个新函数: cout << std::setw(width[j]);    这个是设置输出宽度的,但是默认是在右侧补充空格 所以就要用cout.setf(ios::left);来设置一下左对齐. #include <iostream> #include <cstdio> #include <sstream> #include <vector> #include <string&g…

题目链接:1484 - Alice and Bob's Trip 题意:BOB和ALICE这对狗男女在一颗树上走,BOB先走,BOB要尽量使得总路径权和大,ALICE要小,可是有个条件,就是路径权值总和必须在[L,R]之间,求终于这条路径的权值. 思路:树形dp,dp[u]表示在u结点的权值,往下dfs的时候顺带记录下到根节点的权值总和,然后假设dp[v] + w + sum 在[l,r]内,就是能够的,状态转移方程为 dp[u] = max{dp[v] + w }(bob) dp[u] = m…