主题链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do? problemId=4535 How Many Sets I Time Limit: 2 Seconds      Memory Limit: 65536 KB Give a set S, |S| = n, then how many ordered set group (S1, S2, ..., Sk) satisfies S1 ∩ S2 ∩ ... ∩ Sk = ∅. (Si is…

How Many Sets I Time Limit: 2 Seconds      Memory Limit: 65536 KB Give a set S, |S| = n, then how many ordered set group (S1, S2, ..., Sk) satisfies S1 ∩ S2 ∩ ... ∩ Sk = ∅. (Si is a subset of S, (1 <= i <= k)) Input The input contains multiple cases…

How Many Sets I Time Limit: 2 Seconds      Memory Limit: 65536 KB Give a set S, |S| = n, then how many ordered set group (S1, S2, ..., Sk) satisfies S1 ∩ S2 ∩ ... ∩ Sk = ∅. (Si is a subset of S, (1 <= i <= k)) Input The input contains multiple cases…

How Many Sets II Time Limit: 2 Seconds      Memory Limit: 65536 KB Given a set S = {1, 2, ..., n}, number m and p, your job is to count how many set T satisfies the following condition: T is a subset of S |T| = m T does not contain continuous numbers…

How Many Sets II Time Limit: 2 Seconds      Memory Limit: 65536 KB Given a set S = {1, 2, ..., n}, number m and p, your job is to count how many set T satisfies the following condition: T is a subset of S |T| = m T does not contain continuous numbers…

终于做出来了,激动.... 这道题隐藏得深啊,但若推导下来,就变简单了. 首先,一个集合的子集的个数为2^n=s.注意了,题目求的是有序集合组,并且每个集合是可以重复使用的,怎么办呢?这就要想到多重集合的排列问题了. 一个多重集合有k种元素,每种元素可以无限次使用,求r-排列个数.答案为 k^r个. 这样,我们使用容斥原理: 总个数为(2^n)^k个 包含一个元素的集合有序组为 C(n,1)(2^(n-1))^k 两个的为.....C(n,2)(2^(n-2))^k .... 于是二项式定理+容…

最近做了不少的组合数的题这里简单总结一下下 1.n,m很大p很小 且p为素数p要1e7以下的 可以接受On的时间和空间然后预处理阶乘 Lucas定理来做以下是代码 /*Hdu3037 Saving Beans*/ #include<cstdio> #include<cstring> #include<iostream> #define ll long long #define maxn 1000010 using namespace std; ll T,n,m,p,f[…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3556 How Many Sets I Time Limit: 2 Seconds      Memory Limit: 65536 KB Give a set S, |S| = n, then how many ordered set group (S1, S2, ..., Sk) satisfies S1 ∩ S2 ∩ ... ∩ Sk = ∅. (Si is…

这题被出题人给活活坑了,题目居然理解错了..哎,不想多说. 题意:给两组数,A组为幸运基数,B组为不幸运的基数,问在[low,high]区间内有多少个数:至少被A组中一个数整除,并且不被B中任意一个数整除.|A|<=15. 分析:看到A长度这么小,以及求区间内满足条件的个数问题,容易想到容斥原理,因为不被B中任意一个数整除,所以将B数组所有数取一个最小公倍数LCM,那么就变成了幸运数字都不会被这个LCM整除. 然后枚举子集,实现要将A中元素去除相互整除的情况,比如A = [2,4],这时因为被至…

一道纯粹的容斥原理题!!不过有一个trick,就是会出现重复的,害我WA了几次!! 代码: #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<set> #include<vector> #define ll long long #define mod 55566677 usin…