Time limit  1000 ms Memory limit  131072 kB The National Intelligence Council of X Nation receives a piece of credible information that Nation Y will send spies to steal Nation X’s confidential paper. So the commander of The National Intelligence Cou…

$map$,简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #include<string> #include<iostream> using namespace std; int n,m,k; ]; int main() { while(~sc…

Sona Time Limit:5000MS     Memory Limit:65535KB     64bit IO Format: Submit Status Practice NBUT 1457 Appoint description:  Description Sona, Maven of the Strings. Of cause, she can play the zither. Sona can't speak but she can make fancy music. Her…

UVA - 1025 A Spy in the Metro Secret agent Maria was sent to Algorithms City to carry out an especially dangerous mission. After several thrilling events we find her in the first station of Algorithms City Metro, examining the time table. The Algorit…

 NBUT 1107  盒子游戏 Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:  Practice  Appoint description:  System Crawler  (Aug 13, 2016 10:35:29 PM) Description 有两个相同的盒子,其中一个装了n个球,另一个装了一个球.Alice和Bob发明了一个游戏,规则如下:Alice和Bob轮流操作,Alice先操作每次操作时,游戏者…

NBUT 1105  多连块拼图 Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:  Practice  Appoint description:  System Crawler  (Aug 12, 2016 9:32:14 AM) Description 多连块是指由多个等大正方形边与边连接而成的平面连通图形. -- 维基百科 给一个大多连块和小多连块,你的任务是判断大多连块是否可以由两个这样的小多连块拼成.小多连块…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't u…

在实际的工作中,经常碰到只需要mock一个类的一部分方法,这时候可以用spy来实现. 被测类: public class EmployeeService { public boolean exist(String userName) { checkPrivateExist(userName); checkPublicExist(userName); return true; } private void checkPrivateExist(String userName) { throw new…

http://lightoj.com/volume_showproblem.php?problem=1220 题目大意: 给你一个x,求出满足 x=b^p, p最大是几. 分析:x=p1^a1*p2^a2*...*pn^an; p最大是gcd(a1,a2,...,an). ///他该诉你x,b,p都是整数,所以x,b有可能是负数.当x是负数时,p不能是偶数. #include<stdio.h> #include<string.h> #include<stdlib.h>…

jasmine提供了很多些很实用的处理Promises的方法,首先我们来考虑下面的这个例子: angular.module("myApp.store").controller("StoresCtrl", function($scope, StoreService, Contact) { StoreService.listStores().then(function(branches) { Contact.retrieveContactInfo().then(func…