TrickGCD Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2649    Accepted Submission(s): 1007 Problem Description You are given an array A , and Zhu wants to know there are how many different…

http://acm.hdu.edu.cn/showproblem.php?pid=6053 题意:给定一个数组,我们定义一个新的数组b满足bi<ai 求满足gcd(b1,b2....bn)>=2的数组b的个数 题解:利用容斥定理.我们先定义一个集合f(x)表示gcd(b1,b2...bn)为x倍数的个数(x为质数),我们在定义一个数mi为数组中的最小值,那么集合{f(2)Uf(3)....f(n)}就是我们想要的答案.f(x)=(a1/x)*(a2/x)*.....(ai/x),直接累加肯定…

TrickGCD Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Problem Description You are given an array A , and Zhu wants to know there are how many different array B satisfy the following conditions? * 1≤Bi≤Ai* For…

题目链接 Problem Description You are given an array A , and Zhu wants to know there are how many different array B satisfy the following conditions? 1≤Bi≤Ai For each pair( l , r ) (1≤l≤r≤n) , gcd(bl,bl+1...br)≥2 Input The first line is an integer T(1≤T≤1…

题意:给出n个数$a[i]$,每个数可以变成不大于它的数,现问所有数的gcd大于1的方案数.其中$(n,a[i]<=1e5)$ 思路:鉴于a[i]不大,可以想到枚举gcd的值.考虑一个$gcd(a_1,a_2,a_3…a_n)=d$,显然每个$a_i$的倍数都满足,有$\frac{a_i}{d}$种方案 那么一个d对答案的贡献为\[\prod_{i=1}^{min(a)}{\lfloor\frac{a_i}{d}\rfloor}    \] 但是所有的d计入会有重复情况,考虑容斥,对d进行素数分…

2017 多校2 hdu 6053 TrickGCD 题目: You are given an array \(A\) , and Zhu wants to know there are how many different array \(B\) satisfy the following conditions? \(1≤B_i≤A_i\) For each pair(\(l , r) (1≤l≤r≤n) , gcd(bl,bl+1...br)≥2\) Input The first line…

/* HDU 6053 - TrickGCD [ 莫比乌斯函数,筛法分块 ] | 2017 Multi-University Training Contest 2 题意: 给出数列 A[N],问满足: 1 <= B[i] <= A[i] ; 对任意(l, r) (1<=l<=r<=n) ,使得 gcd(bl,...br) >= 2 ; 的 B[N] 数列的个数 分析: 设 gcd(b1,...bn) = k (k >= 2),此时 k 对答案的贡献为 (a1/k)…

TrickGCD Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 3401    Accepted Submission(s): 1268 Problem Description You are given an array A , and Zhu wants to know there are how many different…

http://acm.hdu.edu.cn/showproblem.php?pid=6053 题意:给出一个A数组,B数组满足Bi<=Ai. 现在要使得这个B数组的GCD值>=2,求共有多少种情况. 思路:在比赛的时候筛了素数表,然后枚举GCD来做,但是还是有些重复情况无办法剔除,赛后才知道是要用莫比乌斯来处理的,然后就趁机学习了一下莫比乌斯. 先是枚举GCD,每个数的可选情况就是GCD/a[i],在这里我们可以把GCD/a[i]相同的数一起处理,也就是用快速幂来计算,这样会显得更快. 那么莫…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=6053 [题目大意] 给出一个数列每个位置可以取到的最大值, 问这个可以构造多少个数列,使得他们的最大公约数大于1 [题解] 我们可以枚举最大公约数k,对于k来说, 他对答案的贡献为∏[ai/k],我们将数列中的数字转化为权值数组 ∏_{i=1}^{100000}[i/k],对于求解i/k的部分我们可以进行数值分块, j*k-1~j*k+k-1的数值除k得到的结果都是相同的,因此可以直接求这个结果…