https://atcoder.jp/contests/abc142/tasks/abc142_d 题意 求满足互素条件下的A和B的因子最多有几个 思路: 分解gcd(A,B)的质因子,再加上1: #include <iostream> #include<algorithm> #include<string> using namespace std; ; long long gcd(long long x,long long y) { )return x; return…

Common Divisors CodeForces - 182D 思路:用kmp求next数组的方法求出两个字符串的最小循环节长度(http://blog.csdn.net/acraz/article/details/47663477,http://www.cnblogs.com/chenxiwenruo/p/3546457.html),然后取出最小循环节,如果最小循环节不相同答案就是0,否则求出各个字符串含有的最小循环节的数量,求这两个数量的公因数个数(也就是最大公因数的因子个数)就是答案.…

C.Common Divisors time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output You are given an array…

You are given an array aa consisting of nn integers. Your task is to say the number of such positive integers xx such that xx divides eachnumber from the array. In other words, you have to find the number of common divisors of all elements in the arr…

Common Divisors time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an array aa consisting of nn integers. Your task is to say the number of such positive integers xx such that x…

B Equal Rectangles 题意: 给你4*n个数,让你判断能不能用这个4*n个数为边凑成n个矩形,使的每个矩形面积相等 题解: 原本是想着用二分来找出来那个最终的面积,但是仔细想一想,那个面积只能是给出的4*n个数中的最小值和最大值的乘积,如果这两个长度不凑成一个矩形,那么肯定全部矩形的面积会出现不一致的 代码: 1 //The idea was to use dichotomies to find that area, and then use that area to figur…

一.题目描述 A common divisor for two positive numbers is a number which both numbers are divisible by. It's easy to calculate the greatest common divisor between tow numbers. But your teacher wants to give you a harder task, in this task you have to find…

题目大意 如果把字符串a重复m次可以得到字符串b,那么我们称字符串a为字符串b的一个因子,现在给定两个字符串S1和S2,求它们的公共因子个数 题解 如果它们有公共因子,那么显然它们的最小公共因子肯定是相等的~~~,公因子就是字符串的最短循环节~~~所以我们先把两个最短循环节给求出来,并判断是否相同,如果相同的话就是它们的最小公因子,然后所有的最小公因子的倍数并且是S1和S2的公约数都是它们的公因子 代码: #include <iostream> #include <cstring>…

http://codeforces.com/problemset/problem/182/D 题意:如果把字符串a重复m次可以得到字符串b,那么我们称字符串a为字符串b的一个因子,现在给定两个字符串S1和S2,求它们的公共因子个数. 思路: 先求最小循环节,如果最小循环节不同,那么肯定是没有公共因子的.如果相同的话,那就看循环节长度为1,2,3...是否可行. #include<iostream> #include<cstdio> #include<cstring> u…

题意: 给你n个数,让你找出来公因子有多少个.公因子:对于这n个数,都能被这个公因子整除 题解: 只需要找出来这n个数的最大公因子x,然后找出来有多少不同数能把x给整.(因为我们可以保证x可以把这n个数整除,又因为x是最大公因数,那么能把x整除的数肯定也可以把这n个数整除) 代码: 1 #include<stdio.h> 2 #include<string.h> 3 #include<iostream> 4 #include<algorithm> 5 #in…

D - Disjoint Set of Common Divisors Problem Statement Given are positive integers AA and BB. Let us choose some number of positive common divisors of AA and BB. Here, any two of the chosen divisors must be coprime. At most, how many divisors can we c…

本文是gtest高级测试指南的译文,由于文章太长,分上下两部分. 一.简介 本文档将向您展示更多的断言,以及如何构造复杂的失败消息,传播致命的故障,重用和加速您的测试夹具,并在您的测试使用各种标志. 二.更多断言 本节包括一些不太常用,但仍然重要的断言. 2.1 显式成功和失败 这三个断言实际上不测试值或表达式. 相反,它们直接产生成功或失败. 与实际执行测试的宏类似,您可以将自定义失败消息流入它们. SUCCEED(); 生成成功. 这不会使整体测试成功. 只有当测试在其执行期间没有任何断言失…

KMP小结   By Wine93 2013.9 1.学习链接: http://www.matrix67.com/blog/archives/115 2.个人小结 1.KMP在字符串中匹配中起着巨大作用,可以在O(n+m)内完成 2.要充分理解next 数组,其求法和next数组的含义(最长后缀等于最长前缀),了解其用途,下面我就next数组在求字符串最小周期中的应用举例 (1).什么是字符串最小周期? Ex:字符串ababab,最小周期为2,为ab Ex:字符串abcd,最小周期为4,为abc…

Codeforces Round #117 (Div. 2) 代码 Codeforces Round #117 (Div. 2) A. Battlefield any trench in meters numerically does not exceed b. 这个条件意味着每次都是在蓄能开始时走向下一条线段,也就是说每条线段相当于花费了\(a+b\)的时间. bfs,用\(d_i\)表示到达线段i需要经过最少的线段条数,到达\(B\)的时候直接计算欧几里得距离. B. Vasya's Cal…

Have you ever played DOTA? If so, you may know the hero, Invoker. As one of the few intelligence carries, Invoker has 10 powerful abilities. One of them is the Ice Wall: Invoker generates a wall of solid ice directly in front of him, and the bitter c…

Chapter 1: Building Abstractions with Procedures 2015-09-29 016 Preface of this chapter QUOTE: The acts of the mind, where in it exerts its power over simple ideas, are chiefly these three ...... (John Locke, An Essay Concerning Human Understanding).…

Machine learning is a branch in computer science that studies the design of algorithms that can learn. Typical machine learning tasks are concept learning, function learning or “predictive modeling”, clustering and finding predictive patterns. These…

The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going to arrange the order of the problems. As we know, the arrangement will have a great effect on the result of the contest. For example, it will…

A. Fraction time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction  is called prope…

B - Problem Arrangement Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3777 Description The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going…

题目:Mike and gcd problem 题意:给一个序列a1到an ,如果gcd(a1,a2,...an)≠1,给一种操作,可以使ai和ai+1分别变为(ai+ai+1)和(ai-ai+1);问需要执行几次这个操作才能使得gcd(a1,a2,...an)>1. 分析: 1.首先,答案总是YES. 2,假设gcd(a1,a2,...an)=1,在一次对ai和ai+1的操作后新的gcd为d,则d满足:d|ai - ai + 1 and d|ai + ai + 1  d|2ai and d|2…

Description Let's call the following process a transformation of a sequence of length nn . If the sequence is empty, the process ends. Otherwise, append the greatest common divisor (GCD) of all the elements of the sequence to the result and remove on…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3777 Time Limit: 2 Seconds      Memory Limit: 65536 KB The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going to arrange the order…

此篇主要了解一下GoogleTest中的断言. 总的来说,GoogleTest中的断言分为两大类:EXPECT_*和ASSERT_*,这两者在测试成功或失败后均会给出测试报告,区别是前者在测试失败后会继续执行下面的测试,而后者在测试失败后会立即终止测试. GoogleTest中的比较断言,涉及整型,字符串, 浮点型,布尔型的比较判断 Fatal assertion Nonfatal assertion Verifies ASSERT_TRUE(condition) EXPECT_TRUE(con…

理解: 本节主要介绍CGAL的代数结构和概念之间的互操作.与传统数论不同,CGAL的代数结构关注于实数轴的“可嵌入”特征.它没有将所有传统数的集合映射到自己的代数结构概念中,避免使用“数的类型”这一术语,由整数入手,逐步提炼,由简入繁,形成一个概念体系.最上层结构是不考虑除法的整数(IntegralDomainWithoutDivision ),它适应了所有“类型都可以由整数生成”的要求.然后向下精炼形成IntegralDomain(这一概念可能就引入了除法).IntegralDomain向下分…

There are less than 60 years left till the 900-th birthday anniversary of a famous Italian mathematician Leonardo Fibonacci. Of course, such important anniversary needs much preparations. Dima is sure that it'll be great to learn to solve the followi…

这几天不知道写点什么,状态也不太好,搬个题上来吧 题意:给定一个数n,设一个从1到n的序列,每次删掉一个序列中的数,求按字典序最大化的GCD序列 做法:按2的倍数找,但是如果除2能得到3的这种情况要特殊处理(￥#……%￥……@#￥不知道该怎么描述,看代码吧) C. Sequence Transformation time limit per test 2 seconds memory limit per test 256 megabytes input standard input output…

520B 给定初始n和目标m,存在两种操作\(-1\)和\(×2\),要求最少操作次数 无脑解法是BFS,不能解决稍大的规模 当n>m时,输出n-m 否则逆向处理,转换为m到n的最小操作次数,存在两种操作\(+1\)和\(/2\)(后者只能在偶数时操作) 由\((m+1+1)/2=m/2+1\)得尽量多的/2操作可以减少操作次数 因此若操作当前的\(m​\)是偶数时尽量/2,直到\(m≤n​\)时再多操作\((n-m)​\)次\(+1​\)是最优操作 Challenge: suppose we…

One of the tasks students routinely carry out in their mathematics classes is to solve a polynomial equation. It is, given a polynomial, sayX2 - 4X + 1<tex2html_verbatim_mark> , to find its roots (2±)<tex2html_verbatim_mark> . If the students'…

Description The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going to arrange the order of the problems. As we know, the arrangement will have a great effect on the result of the contest. For examp…