http://acm.hdu.edu.cn/showproblem.php?pid=6314 题意 对于n*m的方格,每个格子只能涂两种颜色,问至少有A列和B行都为黑色的方案数是多少. 分析 参考https://blog.csdn.net/IcePrincess_1968/article/details/81255138 重点在于计算容斥系数. #include <iostream> #include <cstdio> #include <cstdlib> #inclu…

非常明显的摆了一个NTT模数.... 题目中求恰好\(k\),那么考虑求至少\(k\) 记\(g(k)\)表示至少\(k\)中颜色出现了恰好\(S\)次 那么,\[g(k) = \binom{M}{k} \frac{N!}{(S!)^k (N-Sk)!} * (M-k)^{N-Sk}\] 根据广义容斥原理,记\(f(i)\)表示恰好\(k\)种颜色出现了恰好\(k\)次 那么,\[f(i) = \sum \limits_{k = i}^M (-1)^{k - i} \binom{k}{i} g(…

HDU 4920 Matrix multiplication 题目链接 题意:给定两个矩阵,求这两个矩阵相乘mod 3 思路:没什么好的想法,就把0的位置不考虑.结果就过了.然后看了官方题解,上面是用了bitset这个东西,能够用来存大的二进制数,那么对于行列相乘.事实上就几种情况,遇到0都是0了,1 1得1,2 1,1 2得2,2 2得1.所以仅仅要存下行列1和2存不存在分别表示的二进制数.然后取且bitcount一下的个数,就能够计算出对应的数值了 代码: 暴力: #include <cst…

HDU 2686 Matrix 题目链接 3376 Matrix Again 题目链接 题意:这两题是一样的,仅仅是数据范围不一样,都是一个矩阵,从左上角走到右下角在从右下角走到左上角能得到最大价值 思路:拆点.建图,然后跑费用流就可以,只是HDU3376这题,极限情况是300W条边,然后卡时间过了2333 代码: #include <cstdio> #include <cstring> #include <vector> #include <queue>…

Samwell Tarly is learning to draw a magical matrix to protect himself from the White Walkers. the magical matrix is a matrix with n rows and m columns, and every single block should be painted either black or white. Sam wants to know how many ways to…

Eddy's爱好 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2204 Description Ignatius 喜欢收集蝴蝶标本和邮票,但是Eddy的爱好很特别,他对数字比较感兴趣,他曾经一度沉迷于素数,而现在他对于一些新的特殊数比较有兴趣. 这些特殊数是这样的:这些数都能表示成M^K,M和K是正整数且K>1. 正当他再度沉迷的…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2686 Yifenfei very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.Every time yifenfei should to do is that choose a detour which frome the top left…

matrix Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5569 Description Given a matrix with n rows and m columns ( n+m is an odd number ), at first , you begin with the number at top-left corner (1,1) and you wa…

题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1695 看了别人的方法才会做 参考博客http://blog.csdn.net/shiren_Bod/article/details/5787722 题意 a,b,c,d,k五个数,a与c可看做恒为1,求在a到b中选一个数x,c到d中选一个数y,使得gcd(x,y)等于k,求x和y有多少对. 首先可以想到选取的必是k的倍数,假设是x和y倍,则x和y一定是互质的在,那么就变成了求1到b/k和1到d/k的之…

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…

Matrix 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5671 Description There is a matrix M that has n rows and m columns (1≤n≤1000,1≤m≤1000).Then we perform q(1≤q≤100,000) operations: 1 x y: Swap row x and row y (1≤x,y≤n); 2 x y: Swap column x and c…

原题地址:http://acm.hdu.edu.cn/showproblem.php?pid=5015 解题思路:一看到题目,感觉是杨辉三角形,然后用组合数学做,不过没想出来怎么做,后来看数据+递推思想,感觉可能是矩阵快速幂,可惜一直不知道a*10+3的 +3怎么处理,果然还是图样图森破啊!如果矩阵能搞出来的话,后面的就简单了,真可惜一直到比赛结束也没想出来,看来这种矩阵的题目做的太少了,真后悔线性代数没有认真学.. 今天晚上又想了一会,完全可以把+3那个放到新的一阶矩阵上,值始终等于3,那么对…

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 Swapping II Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1430    Accepted Submission(s): 950 Problem Description Given an N * M matrix with each entry equal to 0 or 1. We can find som…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the product of them. bobo hates big integers. So you are only asked to find t…

要注意取出来的时候 先取出q的是后面那个矩阵 后取出p的是前面的矩阵 所以是判断 p.a == q.b #include <iostream> #include <stack> #include <cstring> #include <cstdio> using namespace std; struct Matrix{ int a,b; Matrix(,):a(aa),b(bb){} }m[]; stack<Matrix>s; int main…

题意: 给一个数的序列,询问一些区间,问区间内与区间其他所有的数都互质的数有多少个. 解法: 直接搞有点难, 所谓正难则反,我们求区间内与其他随便某个数不互质的数有多少个,然后区间长度减去它就是答案了. 那么怎么求区间内与区间其他某个数不互质的数的个数(记为cnt)呢? 我们用L[i],R[i]表示在整个序列中左边与 i 最近的与 i 不互质的数的位置,R[i]表示右边的,L[i],R[i]我们可以正反扫一遍顺便分解因子,用个pos[]记录很方便地求出.那么区间内的cnt为L[i]或R[i]在区…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4920 解题报告:求两个800*800的矩阵的乘法. 参考这篇论文:http://wenku.baidu.com/link?url=261XeEzH-AZkFGPiN63t1nnojoQF50yiuMoviHroGjVXjjRlxFcvWLcws0jgQcmZo4oA9BJcjnPxVreWRu-XXa9zb6r5gUUTxmBXn_qWSsu&qq-pf-to=pcqq.group 我看过了,只是简…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1775    Accepted Submission(s): 796 Problem Description Given two matrices A and B of size n×n, find the product of them.…

Matrix Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 502    Accepted Submission(s): 215 Problem Description There is a matrix M that has n rows and m columns (1≤n≤1000,1≤m≤1000).Then we perf…

Number Sequence Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description Given a number sequence b1,b2…bn. Please count how many number sequences a1,a2,...,an satisfy the condition that a1*a2*...*an=b1*…

题目链接 题意 : 从[a,b]中找一个x,[c,d]中找一个y,要求GCD(x,y)= k.求满足这样条件的(x,y)的对数.(3,5)和(5,3)视为一组样例 . 思路 :要求满足GCD(x,y)=k的对数,则将b/k,d/k,然后求GCD(x,y)=1的对数即可.假设b/k >= d/k ;对于1到b/k中的某个数s,如果s<=d/k,则因为会有(x,y)和(y,x)这种会重复的情况,所以这时候的对数就是比s小的与s互质的数的个数,即s的欧拉函数.至于重复的情况是指:在d/k中可能有大于…

读完题目就知道要使用容斥原理做! 下面用的是二进制实现的容斥原理,详见:http://www.cnblogs.com/xin-hua/p/3213050.html 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector>…

看懂题目,很容易想到容斥原理. 刚开始我用的是二进制表示法实现容斥原理,但是一直超时.后来改为dfs就过了…… 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #defin…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=5213 Lucky Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 763    Accepted Submission(s): 249 Problem Description WLD is always very lucky.His secret…

Problem Description Given a matrix with n rows and m columns ( n+m ,) and you want to go to the number at bottom-right corner (n,m). And you must go right or go down every steps. Let the numbers you go through become an array a1,a2,...,a2k. The cost…

Problem Description Given an N * M matrix with each entry equal to 0 or 1. We can find some rectangles in the matrix whose entries are all 1, and we define the maximum area of such rectangle as this matrix's goodness. We can swap any two columns any…

Matrix multiplication                                                                           Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the…

题意:……应该不用我说了,看起来就很容斥原理,很中国剩余定理…… 方法:因为题目中的n最大是15,使用状态压缩可以将所有的组合都举出来,然后再拆开成数组,进行中国剩余定理的运算,中国剩余定理能够求出同时满足余膜条件的最小整数x,x在(1,M)之间由唯一值,M是各个除数的乘积,所有符合条件的解为ans = x+k*M,可以知道在[1,R]这个区间内,有(M+R-x)/ M个k符合条件,然后在运算中为了防止溢出,所以使用了带膜乘法,就是将乘数转化为二进制,通过位移运算符,在中间过程中不断的取膜(看代…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2686 思路:多线程dp,参考51Nod 1084:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1084 注:这道题用滚动数组优化反而WA,压到三维即可 代码: #include <bits/stdc++.h> using namespace std; ][],dp[][][]; int main() { int n;…