题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3577 题意不好理解,给你数字k表示这里车最多同时坐k个人,然后有q个询问,每个询问是每个人的上车和下车时间,每个人按次序上车,问哪些人能上车输出他们的序号. 这题用线段树的成段更新,把每个人的上下车时间看做一个线段,每次上车就把这个区间都加1,但是上车的前提是这个区间上的最大值不超过k.有个坑点就是一个人上下车的时间是左闭右开区间,可以想到要是一个人下车,另一个人上车,这个情况下这个点的大小还是不变…

Fast Arrangement Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3563    Accepted Submission(s): 1024 Problem Description Chinese always have the railway tickets problem because of its' huge amo…

线段树成段更新+区间最值. 注意某人的乘车区间是[a, b-1],因为他在b站就下车了. #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #define lson l, m, rt << 1 #define rson m + 1, r, rt << 1 | 1 #define lc rt << 1 #define rc…

Problem Description Chinese always have the railway tickets problem because of its' huge amount of passangers and stations. Now goverment need you to develop a new tickets query system.One train can just take k passangers. And each passanger can just…

题目链接.hdu 4965 Fast Matrix Calculation 题目大意:给定两个矩阵A,B,分别为N*K和K*N. 矩阵C = A*B 矩阵M=CN∗N 将矩阵M中的全部元素取模6,得到新矩阵M' 计算矩阵M'中全部元素的和 解题思路:由于矩阵C为N*N的矩阵,N最大为1000.就算用高速幂也超时,可是由于C = A*B, 所以CN∗N=ABAB-AB=AC′N∗N−1B,C' = B*A, 为K*K的矩阵,K最大为6.全然能够接受. #include <cstdio> #inc…

HDU 4965 Fast Matrix Calculation 题目链接 矩阵相乘为AxBxAxB...乘nn次.能够变成Ax(BxAxBxA...)xB,中间乘n n - 1次,这样中间的矩阵一个仅仅有6x6.就能够用矩阵高速幂搞了 代码: #include <cstdio> #include <cstring> const int N = 1005; const int M = 10; int n, m; int A[N][M], B[M][N], C[M][M], CC[N…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1227 题意:一维坐标上有n个点,位置已知,选出k(k <= n)个点,使得所有n个点与选定的点中最近的点的距离总和最小,求出最小值. 思路: 将点i的距离记为为dis[i],从i到j选出一点使此段距离和最小,则此点坐标为dis[(i + j) / 2] cost[i][j]为从i到j选出一点距离和的最小值.则求cost[i][j]的代码如下: cost[i][j] = 0; for(int k =…

题目链接:hdu 4965,题目大意:给你一个 n*k 的矩阵 A 和一个 k*n 的矩阵 B,定义矩阵 C= A*B,然后矩阵 M= C^(n*n),矩阵中一切元素皆 mod 6,最后求出 M 中所有元素的和.题意很明确了,便赶紧敲了个矩阵快速幂的模板(因为编程的基本功不够还是调试了很久),然后提交后TLE了,改了下细节,加了各种特技,比如输入优化什么的,还是TLE,没办法,只好搜题解,看了别人的题解后才知道原来 A*B 已经是 n*n 的矩阵了,所以(A*B)n*n 的快速幂里的每个乘法都是…

Fast Food Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2173    Accepted Submission(s): 930 Problem Description The fastfood chain McBurger owns several restaurants along a highway. Recently,…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4965 题意 给出两个矩阵 一个A: n * k 一个B: k * n C = A * B M = (A * B) ^ (n * n) 然后将M中所有的元素对6取余后求和 思路 矩阵结合律.. M = (A * B) * (A * B) * (A * B) * (A * B) * (A * B) * (A * B) * (A * B) * (A * B) -- 其实也等价于 M = A * (B *…

题目链接 题意 : 有n个饭店,要求建k个供应点,要求每个供应点一定要建造在某个饭店的位置上,然后饭店都到最近的供应点拿货,求出所有饭店到最近的供应点的最短距离. 思路 : 一开始没看出来是DP,后来想想就想通了.预处理,如果要在下标为 i 到 j 的饭店建一个供应点,那一定是在下标为(i+j)/2的位置建造的,状态转移方程:dp[i][j] = dp[i-1][k-1]+dis[k][j](i <= k <= j) ,dp[i][j]代表的是在前 j 个饭店中建了 i 个供应点的最小距离.方…

题意: 给出一个\(n \times k\)的矩阵\(A\)和一个\(k \times n\)的矩阵\(B\),其中\(4 \leq N \leq 1000, \, 2 \leq K \leq 6\). 矩阵\(C=A \cdot B\),求矩阵\(C^{N^2}\)的各个元素之和,以上矩阵运算均是在模\(6\)的情况下计算的. 分析: 如果我们直接计算\(A \cdot B\)的话,这个矩阵非常大,不可能进行快速幂计算. 所以要变形一下, \((A \cdot B)^{N^2}=A \cdot…

一种奇葩的写法,纪念一下当时的RE. #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #include <string> #include <queue> #include <stack> #include <vec…

题意: X轴上有N个餐馆.位置分别是D[1]...D[N]. 有K个食物储存点.每一个食物储存点必须和某个餐厅是同一个位置. 计算SUM(Di-(离第i个餐厅最近的储存点位置))的最小值. 1 <= n <= 200, 1 <= k <= 30, k <= n 思路: 第K个储存点的位置如果确定,前K-1个储存点的位置是浮动的.有很多的重复子结构.DP的结构很明显. dp[i][j]:第i个储存点放在第j个餐馆的位置所得到的最小值. 代码: int n,k; int pos[…

KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题    //2019.3.18 POJ 2251 Dungeon Master POJ 3278 Catch That Cow  //4.8 POJ 3279 Fliptile POJ 1426 Find The Multiple  //4.8 POJ 3126 Prime Path POJ 3087 Shuffle…

layout: post title: 「kuangbin带你飞」专题十九 矩阵 author: "luowentaoaa" catalog: true tags: mathjax: true - kuangbin - 矩阵 传送门 A.CodeForces - 450B Jzzhu and Sequences 题意 水题,主要是拿来试试模板 题解 F[i]=f[i-1]-f[i-2] #include<bits/stdc++.h> using namespace std;…

[kuangbin带你飞]专题1-23 专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil Deposit…

专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil DepositsHDU 1495 非常可乐HDU 26…

One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her. Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face.…

http://acm.hdu.edu.cn/showproblem.php?pid=4965 2014 Multi-University Training Contest 9 1006 Fast Matrix Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 238    Accepted Submission(…

thanks to http://stackoverflow.com/questions/2144459/using-scanf-to-accept-user-input and http://stackoverflow.com/questions/456303/how-to-validate-input-using-scanf for the i/o part. thanks to http://www.haodaima.net/art/137347 for the rounding part…

1.Robberies 连接 :http://acm.hdu.edu.cn/showproblem.php?pid=2955     背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和… 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);    正确的方程是:f[j]=max(f[j],f[j-q[i].money]*q[i…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5327 Olympiad Description You are one of the competitors of the Olympiad in numbers. The problem of this year relates to beatiful numbers. One integer is called beautiful if and only if all of its digita…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4198 Quick out of the Harbour Description Captain Clearbeard decided to go to the harbour for a few days so his crew could inspect and repair the ship. Now, a few days later, the pirates are getting land…

http://acm.hdu.edu.cn/showproblem.php?pid=1155 Bungee Jumping Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 811    Accepted Submission(s): 349 Problem Description Once again, James Bond is fle…

The Donkey of Gui Zhou Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4740 Description There was no donkey in the province of Gui Zhou, China. A trouble maker shipped one and put it in the forest which could be…

Average is not Fast Enough! http://acm.hdu.edu.cn/showproblem.php?pid=1036 Problem Description A relay is a race for two or more teams of runners. Each member of a team runs one section of the race. Your task is to help to evaluate the results of a r…

HDU 动态规划(46道题目)倾情奉献~ [只提供思路与状态转移方程] Robberies http://acm.hdu.edu.cn/showproblem.php?pid=2955      背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和… 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);          正确的…

HDU 2389 Rain on your Parade / HUST 1164 4 Rain on your Parade(二分图的最大匹配) Description You're giving a party in the garden of your villa by the sea. The party is a huge success, and everyone is here. It's a warm, sunny evening, and a soothing wind send…

- Eddy's research II Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description As is known, Ackermann function plays an important role in the sphere of theoretical computer science. However, in the other h…