http://acm.timus.ru/problem.aspx?space=1&num=1221 算是枚举的 题目意思是必须划出这样的 11011 10001 00000 10001 11011 注意中间必须是完整的0,不多不少,旁边的1也是如此 不多不少 才可以 题目很简单 预处理出1的个数 以及以当前0为最低顶角的最大正方形的边长 判断一下1的个数是否满足就可 #include <iostream> #include<cstdio> #include<cstri…

1221. Malevich Strikes Back! Time limit: 1.0 secondMemory limit: 64 MB After the greatest success of Malevich's "Black Square" the famous artist decided to create a new masterpiece. He took a large sheet of checked paper and filled some cells wi…

319. Kalevich Strikes Back Time limit per test: 0.5 second(s)Memory limit: 65536 kilobytes input: standardoutput: standard And yet again the Berland community can see that talent is always multi-sided. The talent is always seeking new ways of self-ex…

H. A + B Strikes Back time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientis…

题意:给定N个数a1,a2,a3...aN,现在要求最小的n满足 n!/(a1!*a2!*...*aN!) 是一个正整数的最小的n. 分析:这题的想法很明确,就是分解a1!*a2!*...*aN!,把其分解成质因子相乘的形式,这个都很熟悉了,然后就是对每一个质因子二分搜索出一个数字下界,最后求其中最大的一个数,问题的关键就是如何分解这样一个表达式成一个质因子相乘的形式.使用一个cnt数组来表示每一个数的在乘积中出现的次数,然后从后往前假设一个数出现了k次,那么如果这个数是素数则不用更新,如果一个…

这道题就是求一个大矩形被n个矩形划分成n+1个部分的面积,这些矩形之间不会相交,可能包含.. #include <cstdio> #include <cstring> #include <vector> #include <algorithm> #define maxn 120100 using namespace std; long long s[maxn]; vector<int>g[maxn]; int pre[maxn]; int X[m…

题目大意: n个矩形,将一个大矩形分成 n+1 块.矩形之间不重合,可是包括.求这n+1个矩形的面积 思路分析: 用线段树记录他们之间的父子关系.然后dfs 计算面积. 当给出的矩形上边的时候,就要记录到该矩形的父亲去. #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #define lson num<<1,s,mid #define rs…

题目链接(是的我越来越懒了..) 题目大意及数据范围: 数据范围很大.“最小”二字让我们考虑二分,但是上界...不会爆long long让你写高精吧? 我们可以发现,∑ai一定满足条件,所以上界是1e13.为什么满足呢?因为C(n+m,n)是整数,所以n!| ( (n+m)!/m! ) ,所以满足. 然后问题只剩下怎么满足判断整除了.数很大,肯定要分解.n!的分解容易解决,因为我们只需考虑1e7范围内的质数即可,每次计算效率约为1e7/ln,总的复杂度能接受.所以只需解决各ai!如何分解.显然不…

Quite often the jury of Saratov SU use the problem "Masquerade" in different practice sessions before the contest. This problem is quite easy — all you need is to print the product of two integers which were read from the input stream. As usual,…

http://mamchenkov.net/wordpress/2017/11/09/systemd-strikes-again-unit-var-whatever-mount-is-bound-to-inactive-unit/ My brother is blogging about a really weird issue we had today.  Apparently, the old school approach of working with disk volumes, and…