题意:给你一些钟的时间,只可以往后调, 问最少调的时间总和是多少 题解:因为肯定是调到某个出现过时间的,只要枚举时间,在维护一个前缀和快速计算出时间总和就行了. #include<cstdio> #include<cmath> #include<vector> #include<map> #include<set> #include<algorithm> #define first fi #define second se using…

题意:有n个时钟,只能顺时针拨,问使所有时间相同的最小代价是多少 思路:将时间排序,枚举拨动到每一个点的时间就好了,容易证明最终时间一定是其中之一 #include <iostream> #include <cstdio> #include <fstream> #include <algorithm> #include <cmath> #include <deque> #include <vector> #include…

CF Gym 102028G Shortest Paths on Random Forests 抄题解×1 蒯板子真jir舒服. 构造生成函数,\(F(n)\)表示\(n\)个点的森林数量(本题都用EGF).怎么求呢 \(f(n)=n^{n-2}\)表示\(n\)个点的树数量,根据\(\exp\)定义,\(e^x=\sum_{i=0}^{\infty}\frac{x^i}{i!}\).那么\(F=\exp f\),感性理解就是如果选\(i\)个联通块拼起来就除以\(i!\),很对的样子. 那么期…

A. Nano alarm-clocks Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/A Description An old watchmaker has n stopped nano alarm-clocks numbered with integers from 1 to n. Nano alarm-clocks count time in hours, and…

A. Nano alarm-clocks Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/A Description An old watchmaker has n stopped nano alarm-clocks numbered with integers from 1 to n. Nano alarm-clocks count time in hours, and…

题目:http://codeforces.com/gym/101933/problem/K 其实每个点的颜色只要和父亲不一样即可: 所以至多 i 种颜色就是 \( i * (i-1)^{n-1} \),设为 \( f(i) \),设恰好 i 种颜色为 \( g(i) \) 那么 \( f(i) = \sum\limits_{j=0}^{i} C_{i}^{j} * g(j) \) 二项式反演得到 \( g(i) = \sum\limits_{j=0}^{k} (-1)^{k-j} * C_{k}…

题目: Description standard input/output As most of you know, the Arab Academy for Science and Technology and Maritime Transport in Alexandria, Egypt, hosts the ACPC Headquarters in the Regional Informatics Center (RIC), and it has been supporting our r…

传送门 A. Ariel time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output King Triton really likes watching sport competitions on TV. But much more Triton likes watching live competitions. So Triton de…

传送门 E. Epic Fail of a Genie time limit per test 0.5 seconds memory limit per test 64 megabytes input standard input output standard output Aladdin had found a new shiny lamp and has started polishing it with his hands. Suddenly a mysterious genie app…

题意:龙要制作n个茶,每个茶的配方是一个字符串,两个字符串之间有一个差值,这个差值为两个字符串每个对应字母之间差的绝对值的最大值,求制作所有茶时获得的所有差值中的最大值. 解法:克鲁斯卡尔.将茶的配方作为点,将每两个点之间的差值作为边权,求最小生成树,这棵树中最大的边即为答案. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<s…