A - Complete the Word(暴力) Description ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 26 where each letter of English alphabet appears exactly once.…

A - A codeforces 714A Description Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya! Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also,…

A - ACodeForces 1060A Description Let's call a string a phone number if it has length 11 and fits the pattern "8xxxxxxxxxx", where each "x" is replaced by a digit. For example, "80123456789" and "80000000000" are ph…

A - A  这题很巧妙啊,前两天刚好做过,而且就在博客里  Little C Loves 3 I time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little C loves number «3» very much. He loves all things about it. Now he has a positive in…

A - A CodeForces - 1042A There are nn benches in the Berland Central park. It is known that aiai people are currently sitting on the ii-th bench. Another mm people are coming to the park and each of them is going to have a seat on some bench out of n…

A.codeforces1038A You are given a string ss of length nn, which consists only of the first kk letters of the Latin alphabet. All letters in string ss are uppercase. A subsequence of string ss is a string that can be derived from ss by deleting some o…

A - Turn the Rectangles CodeForces - 1008B There are nn rectangles in a row. You can either turn each rectangle by 9090 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4941 解题报告:给你一个n*m的矩阵,矩阵的一些方格中有水果,每个水果有一个能量值,现在有三种操作,第一种是行交换操作,就是把矩阵的两行进行交换,另一种是列交换操作,注意两种操作都要求行或列至少要有一个水果,第三种操作是查找,询问第A行B列的水果的能量值,如果查询的位置没有水果,则输出0. 因为n和m都很大,达到了2*10^9,但水果最多一共只有10^5个,我的做法是直接用结构体存了之后排序,然后m…

比赛题目链接 题意:有n个人每人拿着一把枪想要杀死n个怪兽,大写字母代表人,小写字母代表怪兽.A只能杀死a,B只能杀死b,如题目中的图所示,枪的弹道不能交叉.人和怪兽的编号分别是1到n,问是否存在能全部杀死的情况,如果存在则输出编号1到n的每个人杀死的怪兽的编号,如果不能输出"Impossible". 题解:贪心,用递归实现,判断相邻的是否能构成一对,优先构成相邻的,如果不能就递归到前面看是否能构成一对即可. #include<cstdio> #include<cst…

题意很容易转化到这样的问题:在一个强连通的有向图D中是否存在这样的集合划分S + T = D,从S到T集合的边权大于从T到S集合的边权. 即D(i, j)  > B(j, i) + D(j, i).或者等价地对任意集合划分:D(i, j) <= B(j, i) + D(j, i)(*). 实际上若存在可行流f,满足:D(i, j) <= f(i, j) <= B(i, j) + D(i, j),则有对于任意割满足式(*),即可以返回"happy". 关于可行流参…