其实这题并不难啊,但是分解因子的细节一定要小心. \(比如样例48,2是因子说明24也是因子,也就是说假如x存在\) \(那么x一定是因子中的最小数乘上最大数\) \(那我们现在去验证x是否存在,先拿x去整除除数表,看看是否所有除数都是x的因子\) \(然后再去判断x的因子个数是不是等于n(确保除数表包含所有因子)\) \(考虑到d_i<=1e6,极端情况下x=1e12(我并不确定这种情况存在)\) \(所以我们不一个一个判断sqrt(x)内的数是否是因子,而是采取短除法\) ll x=zu,n…

divisors 数学 给定\(m\)个不同的正整数\(a_1, a_2,\cdots, a_m\),请对\(0\)到\(m\)每一个\(k\)计算,在区间\([1, n]\)里有多少正整数是\(a\)中恰好\(k\)个数的约数. 极度考验语文能力的题面. 套路一般分解质因数,但是我们发现分解质因数之后统计会很麻烦,又发现\(m\),\(a_i\)的所有约数个数又很小,所以我们索性将\(m\)个数分别都预处理出所有可能的约数分解形式丢进栈,之后直接sort栈,线性统计答案即可. 另外,我们发现\…

  Product of digits  For a given non-negative integer number N , find the minimal natural Q such that the product of all digits of Q is equal N . Input The first line of input contains one positive integer number, which is the number of data sets. Ea…

BZOJ_4459_[Jsoi2013]丢番图_数学+分解质因数 Description 丢番图是亚历山大时期埃及著名的数学家.他是最早研究整数系数不定方程的数学家之一. 为了纪念他,这些方程一般被称作丢番图方程.最著名的丢番图方程之一是x^N+y^n=z^N.费马 提出,对于N>2,x,y,z没有正整数解.这被称为“费马大定理”,它的证明直到最近才被安德 鲁·怀尔斯(AndrewWiles)证明. 考虑如下的丢番图方程: 1/x+1/y=1/n(x,y,n属于N+)              …

D. Soldier and Number Game time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5812 Description In number theory, a prime is a positive integer greater than 1 that has no positive divisors other than 1 and itself. The distance between two positive integers x and y, denoted by…

Counting Divisors Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 3170    Accepted Submission(s): 1184 Problem Description In mathematics, the function d(n) denotes the number of divisors of…

Divisors 题目链接(点击) Your task in this problem is to determine the number of divisors of Cnk. Just for fun -- or do you need any special reason for such a useful computation? Input The input consists of several instances. Each instance consists of a sin…

---恢复内容开始--- We guessed some integer number xx. You are given a list of almost all its divisors. Almost all means that there are all divisors except 11and xx in the list. Your task is to find the minimum possible integer xx that can be the guessed nu…

题意:有\(n\)组数,对于每组数,问是否能找到两个因子\(d_{1},d{2}\),使得\(gcd(d_{1}+d_{2},a_{i}=1)\),如果有,输出它们,否则输出\(-1\). 题解:对于这题,首先我们要推两个gcd的公式: ​ 1) $gcd(a,b)=gcd(a+b,b) $. ​ 2) 若\(gcd(a,c)=1 \ => gcd(a,bc)=gcd(a,b)\). 这两个公式应该都很容易证明. 因此我们推出:若\(gcd(x,y)=1\),则:\(gcd(x+y,xy)=1\…