来自FallDream的博客,未经允许,请勿转载,谢谢. 设d(x)表示x的约数个数,求$\sum_{i=1}^{n}d(i^{3})$ There are 5 Input files. - Input #1: 1≤N≤10000, TL = 1s. - Input #2: 1≤T≤300, 1≤N≤10^8, TL = 20s. - Input #3: 1≤T≤75, 1≤N≤10^9, TL = 20s. - Input #4: 1≤T≤15, 1≤N≤10^10, TL = 20s. -…

分析 首先,STO ywy OTZ,ywy TQL%%%! 说一下这道题用min_25筛怎么做. 容易发现,对于所有质数\(p\),都满足\(f(p)=4\),于是我们就可以直接通过\([1,x]\)内的质数的个数\(h(x)\)来求出\(g(x)=\sum_{i=1}^{x}f(i) \times [i \in prime]\)了,即\(g(x)\)可以等价地表示为\(g(x)=4 \times h(x)\).如何求\(h(x)\)是min_25筛的基本操作就不过多赘述了.而且进一步分析我们可…

DIVCNT2 - Counting Divisors (square) DIVCNT3 - Counting Divisors (cube) 杜教筛 [学习笔记]杜教筛 (其实不算是杜教筛,类似杜教筛的复杂度分析而已) 你要大力推式子: 把约数个数代换了 把2^质因子个数 代换了 构造出卷积,然后大于n^(2/3)还要搞出约数个数的式子和无完全平方数的个数的容斥... .... 然后恭喜你,spoj上过不去... bzoj能过: #include<bits/stdc++.h> #define…

DIVCNT2 - Counting Divisors (square) #sub-linear #dirichlet-generating-function Let \sigma_0(n)σ​0​​(n) be the number of positive divisors of nn. For example, \sigma_0(1) = 1σ​0​​(1)=1, \sigma_0(2) = 2σ​0​​(2)=2 and \sigma_0(6) = 4σ​0​​(6)=4. LetS_2(…

题目 vjudge URL:Counting Divisors (square) Let σ0(n)\sigma_0(n)σ0​(n) be the number of positive divisors of nnn. For example, σ0(1)=1\sigma_0(1) = 1σ0​(1)=1, σ0(2)=2\sigma_0(2) = 2σ0​(2)=2 and σ0(6)=4\sigma_0(6) = 4σ0​(6)=4. Let S2(n)=∑i=1nσ0(i2).S_2(n…

Counting Divisors Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1604    Accepted Submission(s): 592 Problem Description In mathematics, the function d(n) denotes the number of divisors of p…

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…

Counting Divisors Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Problem Description In mathematics, the function d(n) denotes the number of divisors of positive integer n. For example, d(12)=6 because 1,2,3,4,…

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6069 题目: Counting Divisors Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1235    Accepted Submission(s): 433 Problem Description In mathem…

/** 题目:hdu6069 Counting Divisors 链接:http://acm.hdu.edu.cn/showproblem.php?pid=6069 题意:求[l,r]内所有数的k次方的约数个数之和. 思路: 用(1+e1)*(1+e2)*...*(1+en)的公式计算约数个数. 素数筛出[l,r]内的素因子,然后直接计算结果.(一开始我用vector存起来,之后再处理,结果超时, 时间卡的很紧的时候,vector也会很占用时间.) */ #include<iostream>…