题意: 给出一个矩阵,初始每个位置上的值都为0,然后有两种操作 一种是更改某个位置上的值 另一种是求某个位置附近曼哈顿距离不大于K的所有位置的值的总和 技巧: 坐标旋转,使得操作之后菱形变成方方正正的矩形,(即“曼哈顿距离”转化为“切比雪夫距离”)方便使用树状数组进行计算. 利用哈希进行离散,节约空间,即不开不必要的空间. 坐标旋转: X=x-y Y=x+y 这个结果显示,A’B’C’D’(X,Y)是ABCD(x,y)绕原点(0,0)左旋转45°后的结果,同时长度变为原来的sqrt(2)倍. 由

墨墨购买了一套N支彩色画笔(其中有些颜色可能相同),摆成一排,你需要回答墨墨的提问.墨墨会像你发布如下指令: 1. Q L R代表询问你从第L支画笔到第R支画笔中共有几种不同颜色的画笔. 2. R P Col 把第P支画笔替换为颜色Col.为了满足墨墨的要求,你知道你需要干什么了吗? Input 第1行两个整数N,M,分别代表初始画笔的数量以及墨墨会做的事情的个数.第2行N个整数,分别代表初始画笔排中第i支画笔的颜色.第3行到第2+M行,每行分别代表墨墨会做的一件事情,格式见题干部分. Outp

题意:... 析:可以直接用数状数组进行模拟,也可以用线段树. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #inc

题意:对于两个区间,[si,ei] 和 [sj,ej],若 si <= sj and ei >= ej and ei - si > ej - sj 则说明区间 [si,ei] 比 [sj,ej] 强.对于每个区间,求出比它强的区间的个数. 解题思路:先将每个区间按 e 降序排列,在按 s 升序排列.则对于每个区间而言,比它强的区间的区间一定位于它的前面. 利用数状数组求每个区间[si,ei]前面 满足条件的区间[sj,ej]个数(条件:ej<=ei),再减去前面的和它相同的区间的个

Problem Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them.She finds that trees maybe disharmony and the Disharmony Value between two trees is associated with

wmz的数数(数状数组) 题目描述 \(wmz\)从小就显现出了过人的天赋,他出生的第三天就证明了哥德巴赫猜想,第五天就证明了质能方程,出生一星期之后,他觉得\(P\)是否等于\(NP\)这个问题比前面他证明的这些定理好玩多了,于是他成为了一名计算机科学家. 在他开始接触计算机科学的第一天,他就已经刷遍了所有\(oj\),这为他今后建立王国推行全民刷题计划打下了坚实的基础.在他接触计算机科学一星期之后,他就已经通读了所有顶级会议的\(paper\),并且在多方面同时出了许多成果. 他对他王国的国

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. Input The firs

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18543    Accepted Submission(s): 11246 Problem Description The inversion number of a given number sequence a1, a2, ..., a

Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2611    Accepted Submission(s): 973 Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the str

Ultra-QuickSort Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 70674   Accepted: 26538 Description In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swappin

Given a list of N integers A1, A2, A3,...AN, there's a famous problem to count the number of inversions in it. An inversion is defined as a pair of indices i < j such that Ai > Aj. Now we have a new challenging problem. You are supposed to count the

#include<bits/stdc++.h> using namespace std; ; int n; int c[maxn]; int lowbit(int x) { return x&(-x); } void update(int num,int val) { while(num) { c[num]+=val; num-=lowbit(num); } } int getsum(int num) { ; while(num<=n) { sum+=c[num]; num+=l

You are given an N × N matrix. At the beginning every element is 0. Write a program supporting 2 operations: 1. Add x y value: Add value to the element Axy. (Subscripts starts from 0 2. Sum x1 y1 x2 y2: Return the sum of every element Axy for x1 ≤ x

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5869 Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 This is a simple problem. The teacher gives Bob a list of problems about GCD (Great

题意:      给你一些人,每个人有自己的攻击力,输入的顺序就是每个人的顺序,他们之间互相比赛,两个人比赛的条件是必须在他们两个位置之间找到一个人当裁判,这个裁判的攻击力必须在他们两个人之间,问你最多能矩形多少场比赛. 思路:       枚举每一个点当裁判,线段树或者树状数组正着跑一遍记录大于等于当前点和小于等于当前点的个数,然后倒着跑一边,做同样处理,最后乘一下就行了. #include<stdio.h> #include<string.h> __int64 num[1100

敌兵布阵 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 99913    Accepted Submission(s): 42300 Problem Description C国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又开始忙乎了.A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务

题目描述 $liu\_runda$决定提高一下知识水平,于是他去请教郭神.郭神随手就给了$liu\_runda$一道神题,$liu\_runda$并不会做,于是把这个题扔到联考里给高二的做.郭神有$n$条位于第一象限内的线段,给出每条线段与$x$轴和$y$轴交点的坐标,显然这样就可以唯一确定每一条线段.$n$条线段和$y$轴交点的纵坐标分别为$1,2,3,4...n$.我们记和$y$轴交点纵坐标为$i$的线段和$x$轴交点的横坐标为$x_i+1,x_i$按这样的方式生成:$x_1$由输入给出.$

题意: 给N个数.a[1]....a[N]. M种操作: S X Y:令a[X]=Y Q L R D P:查询a[L]...a[R]中满足第D位上数字为P的数的个数 数据范围: 1<=T<= 501<=N, M<=1000000<=a[i]<=$2^{31}$ - 11<=X<=N0<=Y<=$2^{31}$ - 11<=L<=R<=N1<=D<=100<=P<=9 思路: 直接开tree[maxn][1

题目链接 /* 按从左到右,从下到上的顺序给出星星的坐标,计算出level为[0,n)的星星的个数. 星星的level为该星星左下边(包括自己正下方的星星,但是不包括自己)星星的个数. BIT模板题. £:把星星的下标转化成从1开始. £:想到用BIT做. */ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=35000+5; const

Cube Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 1166    Accepted Submission(s): 580 Problem Description Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number

Color the ball Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5397    Accepted Submission(s): 2882 Problem Description N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球