原题链接:http://codeforces.com/gym/100338/attachments/download/2136/20062007-winter-petrozavodsk-camp-andrew-stankevich-contest-22-asc-22-en.pdf 题意 给你n个点,让你连边,使得每个点的度至少为1,问你方案数. 题解 从正面考虑非常困难,应从反面考虑,取 i 点出来不连,这样的取法一共有C(n,i)种取法,其他的连好,而这样就是子问题了.那么dp[n]=2^(n…

递推就好了,用二项式定理算出所有连边的方案数,减去不合法的方案, 每次选出一个孤立点,那么对应方案数就是上次的答案. 枚举选几个孤立点和选哪些,选到n-1个点的时候相当于都不选,只减1. 要用到高精度,直接开100*100的组合数数组会MLE,用滚动数组优化一下就好了. 不会java,python太伤了 #include<bits/stdc++.h> using namespace std; ; struct bign { int len, s[MAXN]; bign () { memset(…

Problem D. Dinner ProblemTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/attachments Description A group of k students from Cooking University living in the campus decided that each day of the semester one of them will p…

Problem F "Folding" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100002 Description Bill is trying to compactly represent sequences of capital alphabetic characters from 'A' to 'Z' by folding repeating subsequences insid…

C. The Intriguing Obsession time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output — This is not playing but duty as allies of justice, Nii-chan! — Not allies but justice itself, Onii-chan! With h…

原题链接:http://codeforces.com/gym/100431/attachments/download/2421/20092010-winter-petrozavodsk-camp-andrew-stankevich-contest-37-asc-37-en.pdf 题意 给你一个n,问你有多少a和x满足:x在a中二分会返回true,其中a的长度是n 题解 考虑到二分的过程不是向左就是向右,所以可以暴力搜索搞到若干序列,这些序列都是由向左或者向右组成的.枚举x,设向左的有i个,向右…

题目链接: B.Table 题意: \(n*m\)的矩阵使每个\(n*n\)矩阵里面准确包含\(k\)个点,问你有多少种放法. \((1 ≤ n ≤ 100; n ≤ m ≤ 10^{18}; 0 ≤ k ≤ n^2)\) 题解: - Let \(s_i\) number of points in the column \(i\). - Two neighboring squares are drawn at this picture, \(A\) is the number of point…

 2017 JUST Programming Contest 2.0 题目链接:Codeforces gym 101343 J.Husam and the Broken Present 2 J. Husam and the Broken Present 2 time limit per test:1.0 s memory limit per test:256 MB input:standard input output:standard output After you helped Husam…

题目链接:http://codeforces.com/contest/219/problem/D 树dp //#pragma comment(linker, "/STACK:102400000, 102400000") #include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include…

一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈算法. 介绍一下floyd判圈算法:该算法适用于在线性时间复杂度内判断有限自动机.迭代函数.链表中是否有环,求环的起点(即链长)和环长. 可以先这么做:首先从起点S出发,给定两个指针,一个快指针一个慢指针,然后每次快指针走1步,慢指针走2步,直到相遇为止.如果已经到达终点/达到规定步数时仍然没有相遇…