题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5742 AC代码: /* * 反思: * 1.遇到简单题别激动,先把它所有的输入任务都完成了,在考虑处理的问题 * 2.当你的程序没有按回车就全部运行完的时候,这段程序十有八九是有问题的, * 仔细检查是不是输入没输入全,或者特殊字符没处理或者运行过程中遇到问题了. */ #include <cstdio> using namespace std; int test…

Matrix Searching Time Limit: 10 Seconds      Memory Limit: 32768 KB Given an n*n matrix A, whose entries Ai,j are integer numbers ( 1 <= i <= n, 1 <= j <= n ). An operation FIND the minimun number in a given ssub-matrix. Input The first line o…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1859 Matrix Searching Time Limit: 10 Seconds      Memory Limit: 32768 KB Given an n*n matrix A, whose entries Ai,j are integer numbers ( 1 <= i <= n, 1 <= j <= n ). An operation FIND t…

Let us consider undirected graph G = which has N vertices and M edges. Incidence matrix of this graph is N * M matrix A = {a ij}, such that a ij is 1 if i-th vertex is one of the ends of j-th edge and 0 in the other case. Your task is to find the sum…

题目来源: 点击打开链接 题目翻译: 矩阵乘法问题是动态规划的典型例子. 假设你必须评估一个表达式,如A * B * C * D * E,其中A,B,C,D和E是矩阵.由于矩阵乘法是关联的,乘法运算的次序是任意的.但是,所需的基本乘法的数量很大程度上取决于您选择的评估顺序. 例如,设A是50 * 10矩阵,B是10 * 20矩阵,C是20 * 5矩阵. 计算A * B * C有两种不同的策略,即(A * B)* C和A *(B * C). 第一个需要15000次基本乘法,但第二个只需要3500次…

Matrix Multiplication Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on ZJU. Original ID: 231664-bit integer IO format: %lld      Java class name: Main   Let us consider undirected graph G = <v, e="">which has N verti…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4836 因为要使对角线所有元素都是U,所以需要保证每行都有一个不同的列上有U,设(i,j)的位置是U, 以U为边,连接点i和点j+n,也即连接行点和列点,最大匹配为n则必定有解,否则必定无解 #include <cstdio> #include <iostream> #include <cstring> #include <cctype>…

题目链接 给一个n*n的矩阵, 给q个查询, 每次给出x1, y1, x2, y2, 求这个矩阵中的最小值. 代码基本上和上一题相同... #include<bits/stdc++.h> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2316 题目意思:有 N 个 点,M 条 边.需要构造一个N * M 大小的矩阵A.对于 (i, j) 这个坐标点它表示,对编号为 i 这个点编号为 j 的 点与它相连,此时标记(i, j) 为1,如果坐标点没有跟这条 j 的边相连,就标记为0.构造完这个矩阵A之后,需要求出它的转置矩阵AT,算出 ATA 的和. 新学期第一场比赛!刚开始真是打算直接做的,但是数据…

第十三届浙江省大学生程序设计竞赛 I 题, 一道模拟题. ZOJ  3944http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3944 In a BG (dinner gathering) for ZJU ICPC team, the coaches wanted to count the number of people present at the BG. They did that by having the waitre…

3861 - Valid Pattern Lock Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3861 Description Pattern lock security is generally used in Android handsets instead of a password. The pattern lock can b…

QS Network Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1586 Appoint description:  System Crawler  (2015-05-31) Description Sunny Cup 2003 - Preliminary Round April 20th, 12:00 - 17:00 Problem…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1094 ZOJ Problem Set - 1094 Matrix Chain Multiplication -------------------------------------------------------------------------------- Time Limit: 2 Seconds Memory Limit: 65536 KB -------…

题目链接 ZOJ链接 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task…

Searching the String Time Limit: 7 Seconds      Memory Limit: 129872 KB Little jay really hates to deal with string. But moondy likes it very much, and she's so mischievous that she often gives jay some dull problems related to string. And one day, m…

Abs Problem Time Limit: 2 Seconds Memory Limit: 65536 KB Special Judge Alice and Bob is playing a game, and this time the game is all about the absolute value! Alice has N different positive integers, and each number is not greater than N. Bob has a…

POJ 1502 MPI Maelstrom / UVA 432 MPI Maelstrom / SCU 1068 MPI Maelstrom / UVALive 5398 MPI Maelstrom /ZOJ 1291 MPI Maelstrom (最短路径) Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shar…

H - K-Nice Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3212 Description This is a super simple problem. The description is simple, the solution is simple. If you believe so, just read it on. O…

Seven Segment Display Time Limit: Seconds Memory Limit: KB A seven segment display, or seven segment indicator, is a form of electronic display device for displaying decimal numerals that is an alternative to the more complex dot matrix displays. Sev…

B - Cryptography Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 2671 Description Young cryptoanalyst Georgie is planning to break the new cipher invented by his friend Andie. To do this, he must…

How Many Shortest Path Time Limit: 10 Seconds      Memory Limit: 32768 KB Given a weighted directed graph, we define the shortest path as the path who has the smallest length among all the path connecting the source vertex to the target vertex. And i…

ZOJ 3690 题意: 有n个人和m个数和一个k,如今每一个人能够选择一个数.假设相邻的两个人选择同样的数.那么这个数要大于k 求选择方案数. 思路: 打表推了非常久的公式都没推出来什么可行解,好不easy有了想法结果WA到天荒地老也无法AC.. 于是学习了下正规的做法,恍然大悟. 这道题应该用递推 + 矩阵高速幂. 我们设F(n) = 有n个人,第n个人选择的数大于k的方案数: G(n) = 有n个人.第n个人选择的数小于等于k的方案数: 那么递推关系式即是: F(1)=m−k,G(1)=k…

题目 给定矩阵A, B和模数p,求最小的正整数x满足 A^x = B(mod p). 分析 与整数的离散对数类似,只不过普通乘法换乘了矩阵乘法. 由于矩阵的求逆麻烦,使用 $A^{km-t} = B(mod \ p)$ 形式的BSGS. 然后就是判断矩阵是否相等, 一种方法是对矩阵进行Hash, 这里为了防止两个不同矩阵的Hash值冲突,使用了两个底数进行Hash. #include<bits/stdc++.h> using namespace std; typedef long long l…

今天我们要讲的是ng2的路由的第二部分,包括路由嵌套.路由生命周期等知识点. 例子 例子仍然是上节课的例子:…

March 16, 2016 Problem statement:Given a 2D array (matrix) named M, print all items of M in a spiral order, clockwise.For example: M  =  1   2   3   4   5       6   7   8   9  10      11  12  13  14  15      16  17  18  19  20 The clockwise spiral pr…

Atitit Data Matrix dm码的原理与特点 Datamatrix原名Datacode,由美国国际资料公司(International Data Matrix, 简称ID Matrix)于1989年发明. Datamatrix是一种矩阵式二维条码,其发展的构想是希望在较小的条码标签上存入更多的资料量.Datamatrix的最小尺寸是目前所有条码中最小的,尤其特别适用于小零件的标识,以及直接印刷在实体上. Datamatrix又可分为ECC000-140与ECC200两种类型,ECC0…

转自:http://www.cnblogs.com/qiengo/archive/2012/06/30/2570874.html#translate Matrix的数学原理 在Android中,如果你用Matrix进行过图像处理,那么一定知道Matrix这个类.Android中的Matrix是一个3 x 3的矩阵,其内容如下: Matrix的对图像的处理可分为四类基本变换: Translate           平移变换 Rotate                旋转变换 Scale    …

Affine Transformation是一种二维坐标到二维坐标之间的线性变换,保持二维图形的"平直性"和"平行性".仿射变换可以通过一系列的原子变换的复合来实现,包括:平移(Translation).缩放(Scale).翻转(Flip).旋转(Rotation)和错切(Shear). 在做2D图形引擎时,仿射变换是非常重要的点,图形的旋转等各种表现都需要通过仿射变换来完成,比如在显示列表树中,父节点旋转了,那么子节点在计算显示时也要叠加上父节点的变换矩阵,这是叠…

Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix. Note that it is the kth smallest element in the sorted order, not the kth distinct element. Example: matrix = [ [ 1, 5…