UVA.11806 Cheerleaders (组合数学 容斥原理 二进制枚举) 题意分析 给出n*m的矩形格子,给出k个点,每个格子里面可以放一个点.现在要求格子的最外围一圈的每行每列,至少要放一个点,并且放在角上的点,同时算那个角所在的行和所在的列.不允许剩下点,求总共的方案数量,结果对1000007取模. 数据范围2 ≤ M,N ≤ 20,K ≤ 500. 考虑到要求组合数目,首先就需要预处理500以内的组合数.正向求解可能有些困难,这样考虑: 不管三七二十一,先求解出所有情况的总和,即C…

HDU.1796 How many integers can you find ( 组合数学 容斥原理 二进制枚举) 题意分析 求在[1,n-1]中,m个整数的倍数共有多少个 与 UVA.10325 The Lottery 一模一样. 前置技能和其一样,但是需要注意的有一下几点: 1. m个数字中可能有0 2. 要用long long 代码总览 #include <cstdio> #include <algorithm> #include <cstring> #incl…

In most professional sporting events, cheerleaders play a major role in entertaining the spectators. Theirroles are substantial during breaks and prior to start of play. The world cup soccer is no exception.Usually the cheerleaders form a group and p…

11806 - Cheerleaders Time limit: 2.000 seconds C Cheerleaders In most professional sporting events, cheerleaders play a major role in entertaining the spectators. Their roles are substantial during breaks and prior to start of play. The world cup soc…

In most professional sporting events, cheerleaders play a major role in entertaining the spectators. Theirroles are substantial during breaks and prior to start of play. The world cup soccer is no exception.Usually the cheerleaders form a group and p…

题目链接:https://www.luogu.org/problemnew/show/UVA11806 容斥原理+组合数 正着找合♂fa的不好找,那就用总方案数-不合♂fa的 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 1000; const int mod = 1e6 +…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5556    Accepted Submission(s): 1593 Problem Description   Now you get a number N, and a M-integers set, you shou…

题意 一个n*m的区域内,放k个啦啦队员,第一行,最后一行,第一列,最后一列一定要放,一共有多少种方法. 思路 设A1表示第一行放,A2表示最后一行放,A3表示第一列放,A4表示最后一列放,则要求|A1∧A2∧A3∧A4| 由容斥原理可知|∪Ai| = Σ|Ai| - Σ|Ai∧Aj| + -- (+-)|Ai∧Aj∧--∧Ak|. 再由德摩根定律得:∧Ai = Cu(∪Cu(Ai)),所以|∧Ai| = S - |∪Cu(Ai)|.(Cu表示集合的非) 然后令A表示不放第一行,B表示不放最后一…

UVA.10325 The Lottery (组合数学 容斥原理) 题意分析 首先给出一个数n,然后给出m个数字(m<=15),在[1-n]之间,依次删除给出m个数字的倍数,求最后在[1-n]之间还剩下多少个数字(包括1和n),已知m个数字中不会包含1(否则全部都被刷掉了). 前置技能 1. 给出数字s,在[1-n]之间,s的倍数有n/s个. 2. 给出数字s1,和s2,在[1-n]之间,既是s1的倍数,又是s2的倍数,有n/lcm(s1,s2)个. 3. 给出数字s1,s2--sk(共k个数字…

题意 PDF 分析 如果要求是某行某列没有石子很好算,就一个组合数. 然后要求某行某列有,就用容斥原理就行了. 时间复杂度\(O(k^2 + 16T)\) 代码 #include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<set> #include<map> #include<queue> #include<stack&…