http://acm.hdu.edu.cn/showproblem.php?pid=2119 Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3097    Accepted Submission(s): 1429 Problem Description Give you a matrix(only contains 0 or 1),ev…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2119 Matrix Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2205    Accepted Submission(s): 975 Problem Description Give you a matrix(only contains…

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…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2119 解题思路: 处理数据,使用公式最小点覆盖数=最大匹配数,使用匈牙利算法求二分图最大匹配即可. AC代码: #include<stdio.h> #include<string.h> ][],book[],cx[],cy[]; int Maxmatch(); int path(int u); int main() { int i,j; ) { scanf("%d"…

Matrix Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3095    Accepted Submission(s): 1428 Problem Description Give you a matrix(only contains 0 or 1),every time you can select a row or a colu…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 820    Accepted Submission(s): 328 Problem Description Given two matrices A and B of size n×n, find the product of them. b…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=5015 由于是个二维的递推式,当时没有想到能够这样构造矩阵.从列上看,当前这一列都是由前一列递推得到.依据这一点来构造矩阵.令b[i]代表第i列,是一个(n+2)*1的矩阵,即b[1] = [1,233......],之所以在加了两行,是要从前一个矩阵b[i-1]得到b[i]中的第二个数2333...,再构造一个转换矩阵a,它是一个(n+2)*(n+2)的矩阵,那么a^(m-1) * b就是第m列. /* a矩…

THE MATRIX PROBLEM Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8693    Accepted Submission(s): 2246 Problem Description You have been given a matrix CN*M, each element E of CN*M is positive…