[BZOJ3120]Line Description Wayne喜欢排队……不对,是Wayne所在学校的校长喜欢看大家排队,尤其是在操场上站方阵.某日课间操时,校长童心大发想了一个极具观赏性的列队方案,如下:1. 方阵排成N行,每行恰好M个学生.2. 由于校长喜欢女孩子,所以在一行上不能有连续P个男生.3. 由于校长喜欢女孩子,所以在校长看来,一列全是男生是不好的,全男生的列数不能超过Q.Wayne因为感冒了所以不用参加列队,不过他看着大家排队排得不亦乐乎,于是他想知道,在男女生数目无限制的情况…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4307    Accepted Submission(s): 2586 Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) =…

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13677   Accepted: 9697 Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequenc…

C. Sasha and Array time limit per test:5 seconds memory limit per test:256 megabytes input:standard input output: standard output Sasha has an array of integers a1, a2, ..., an. You have to perform m queries. There might be queries of two types: 1 l…

对于任意矩阵M和N,若矩阵M的列数等于矩阵N的行数,则记M和N的乘积为P=M*N,其中mik 记做矩阵M的第i行和第k列,nkj记做矩阵N的第k行和第j列,则矩阵P中,第i行第j列的元素可表示为公式(1-1): pij=(M*N)ij=∑miknkj=mi1*n1j+mi2*n2j+--+mik*nkj (公式1-1) 由公式(1-1)可以看出,最后决定pij是(i,j),所以可以将其作为Reducer的输入key值.为了求出pij分别需要知道mik和nkj,对于mik,其所需要的属性有矩阵M,…

hdu4920 Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 568    Accepted Submission(s): 225 Problem Description Given two matrices A and B of size n×n, find the product o…

codevs 1250 Fibonacci数列  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 钻石 Diamond   题目描述 Description 定义:f0=f1=1, fn=fn-1+fn-2(n>=2).{fi}称为Fibonacci数列. 输入n,求fn mod q.其中1<=q<=30000. 输入描述 Input Description 第一行一个数T(1<=T<=10000). 以下T行,每行两个数,n,q(n<=109, …

Cow Relays Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7335   Accepted: 2878 Description For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout…

DNA Sequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14758 Accepted: 5716 Description It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence,For exampl…

So Easy! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3804    Accepted Submission(s): 1251 Problem Description A sequence Sn is defined as: Where a, b, n, m are positive integers.┌x┐is the ce…

Matrix Power Series Time Limit: 3000MS   Memory Limit: 131072K Total Submissions: 12346   Accepted: 5262 Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak. Input The input contains exactly one test cas…

Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 .Now, I will give a0 ~ a9 and two positiv…

先贴两个基础的矩阵乘法,以后再贴一些题. hdu. Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 5236    Accepted Submission(s): 2009 Problem Description Given two matrices A and B of size n×n,…

转载请注明出处. /* Function:C++实现并行矩阵乘法; Time: 19/03/25; Writer:ZhiHong Cc; */ 运行方法:切到工程文件x64\Debug文件下,打开命令行,输入以下指令: mpiexec -n N Project.exe NUM // N代表开启进程数量,NUM代表矩阵规模大小(size) 具体代码: 1.头文件: #include<stdio.h> #include <iostream> #include<math.h>…

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13677   Accepted: 9697 Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequenc…

来自:http://blog.csdn.net/xyilu/article/details/9066973 引言 何 为大矩阵?Excel.SPSS,甚至SAS处理不了或者处理起来非常困难,需要设计巧妙的分布式方法才能高效解决基本运算(如转置.加法.乘法.求逆) 的矩阵,我们认为其可被称为大矩阵.这意味着此种矩阵的维度至少是百万级的.经常是千万级的.有时是亿万级的.举个形象的栗子.至2012年12月底,新 浪微博注册用户数超5亿,日活跃用户4629万[1],如果我们要探索这4000多万用户可以分…

We consider problems concerning the number of ways in which a number can be written as a sum. If the order of the terms in the sum is taken into account the sum is called a composition and the number of compositions of n is denoted by c(n). Thus, the…

▶ 按书上的步骤使用不同的导语优化矩阵乘法 ● 所有的代码 #include <iostream> #include <cstdlib> #include <chrono> #define SIZE 1024 using namespace std; using namespace std::chrono; double a[SIZE][SIZE], b[SIZE][SIZE], c[SIZE][SIZE], d[SIZE][SIZE];// 四个数组放入 main 里…

题目描述 [问题描述] 编写程序,完成3*4矩阵和4*3整数矩阵的乘法,输出结果矩阵. [输入形式] 一行,供24个整数.以先行后列顺序输入第一个矩阵,而后输入第二个矩阵. [输出形式] 先行后列顺序输出结果矩阵,每个元素的显示宽度为8格,屏幕一行只显示矩阵的一行. [样例输入] 1 2 3 4 5 6 7 8 9 1 2 3 9 8 7 6 5 4 3 2 1 1 2 3 上面的输入,意味着要计算如下两个矩阵的乘积. 第一个矩阵 : 1 2 3 4 5 6 7 8 9 1 2 3 第二个矩阵:…

矩阵乘法模板: #define N 801 #include<iostream> using namespace std; #include<cstdio> int a[N][N],b[N][N],c[N][N]; int n,m,p; int read() { ,ff=;char s; s=getchar(); ') { ; s=getchar(); } ') { ans=ans*+s-'; s=getchar(); } return ans*ff; } int main() {…

矩阵乘法来进行所有路径的运算, 线段树来查询修改. 关键还是矩阵乘法的结合律. Harry And Math Teacher Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 326    Accepted Submission(s): 89 Problem Description As we all know, Harry Porter…

Cow Relays Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7825   Accepted: 3068 Description For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout…

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…

题目链接:https://vjudge.net/problem/HDU-4965 Fast Matrix Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2057    Accepted Submission(s): 954 Problem Description One day, Alice and Bob…

E. Xor-sequences time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given n integers a1,  a2,  ...,  an. A sequence of integers x1,  x2,  ...,  xk is called a "xor-sequence" if f…

题目描述 In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field, or, more generally, in a ring or even a semiring. The matrix product is designed for representing the…

Matrix Power Series Time Limit: 3000MS   Memory Limit: 131072K Total Submissions: 23187   Accepted: 9662 Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak. Input The input contains exactly one test cas…

Description 给你一个N*N的矩阵,不用算矩阵乘法,但是每次询问一个子矩形的第K小数. Input 第一行两个数N,Q,表示矩阵大小和询问组数: 接下来N行N列一共N*N个数,表示这个矩阵: 再接下来Q行每行5个数描述一个询问:x1,y1,x2,y2,k表示找到以(x1,y1)为左上角.以(x2,y2)为右下角的子矩形中的第K小数. Output 对于每组询问输出第K小的数. Sample Input 2 2 2 1 3 4 1 2 1 2 1 1 1 2 2 3 Sample Out…

Matrix Power Series Time Limit: 3000MS Memory Limit: 131072K Description Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + - + Ak. Input The input contains exactly one test case. The first line of input contains three po…

Description For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture. Each trail connects two different intersections (1 ≤ I1i ≤ 1,000; 1 ≤ I2i ≤ 1,000…