GCD SUM Time Limit: 8000/4000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description 给出N,M执行如下程序:long long  ans = 0,ansx = 0,ansy = 0;for(int i = 1; i <= N; i ++)   for(int j = 1; j <= M; j ++)     …

[BZOJ2820]YY的GCD(莫比乌斯反演) 题面 讨厌权限题!!!提供洛谷题面 题解 单次询问\(O(n)\)是做过的一模一样的题目 但是现在很显然不行了, 于是继续推 \[ans=\sum_{d=1}^n[d\_is\_prime]\sum_{i=1}^{n/d}[\frac{n}{id}][\frac{m}{id}]\] 老套路了 令\(T=id\) \[ans=\sum_{T=1}^{n}[\frac{n}{T}][\frac{m}{T}]\sum_{d|T}[d\_is\_prim…

[BZOJ2818]Gcd(莫比乌斯反演) 题面 Description 给定整数N,求1<=x,y<=N且Gcd(x,y)为素数的 数对(x,y)有多少对. Input 一个整数N Output 如题 Sample Input 4 Sample Output 4 HINT 对于样例(2,2),(2,4),(3,3),(4,2) 1<=N<=10^7 题解 题目要求的: \[\sum_{i=1}^n\sum_{j=1}^n[gcd(i,j)\_is\_prime]\] 把因数提出来…

[HDU1695]GCD(莫比乌斯反演) 题面 题目大意 求\(a<=x<=b,c<=y<=d\) 且\(gcd(x,y)=k\)的无序数对的个数 其中,你可以假定\(a=c=1\) 所有数都\(<=100000\) 数据组数\(<=3000\) 题解 莫比乌斯反演 作为一道莫比乌斯反演的题目 首先我们要迈出第一步 如果有\(gcd(x,y)=k\) 那么,我们就有\(gcd(\frac{x}{k},\frac{y}{k})=1\) 所以,现在问题相当于转化为了求 \(…

题目:GCD SUM 题目链接:http://www.bnuoj.com/v3/problem_show.php?pid=39872 算法:莫比乌斯反演.优化 #include<stdio.h> #define N 100001 typedef long long LL; }; ; int mu[N]; LL f[N],ff[N]; //缩短时间 /* 莫比乌斯函数mu[i]的定义: 1. 如果 i 是素数,那么mu[i]为-1; 2. 如果 i 是由多个不同的素数组成的,那么mu[i]为-1…

SPOJ Problem Set (classical) 7001. Visible Lattice Points Problem code: VLATTICE Consider a N*N*N lattice. One corner is at (0,0,0) and the opposite one is at (N,N,N). How many lattice points are visible from corner at (0,0,0) ? A point X is visible…

BUPT2017 wintertraining(15) #5H HDU- 4947 题意 有一个长度为l的数组,现在有m个操作,第1种为1 n d v,给下标x 满足gcd(x,n)=d的\(a_x\)增加v.第2种为2 x,查询\(\sum_{i=1}^x a_i\). 数据范围:\(1\le n,d,v\le2\cdot 10^5,1\le x\le l\) 题解 设\(f_i\)满足\(a_i=\sum_{d|i} f_d\),用树状数组存储\(f_i\)的前缀和. \[a_x+=v\cd…

4491. Primes in GCD Table Problem code: PGCD Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has cre…

[题目链接] https://www.luogu.org/problemnew/show/P2257 // luogu-judger-enable-o2 /* ----------------------- [题解] https://www.luogu.org/blog/peng-ym/solution-p2257 [莫比乌斯反演] http://www.cnblogs.com/peng-ym/p/8647856.html [整除分块] http://www.cnblogs.com/peng-y…

点此看题面 大致题意: 求\(\sum_{x=1}^N\sum_{y=1}^MIsPrime(gcd(x,y))\). 莫比乌斯反演 听说此题是莫比乌斯反演入门题? 一些定义 首先,我们可以定义\(f(d)\)和\(F(d)\)如下: \[f(d)=\sum_{i=1}^N\sum_{j=1}^M[gcd(i,j)==d]\] \[F(d)=\sum_{i=1}^N\sum_{j=1}^M[d|gcd(i,j)]\] 通过定义,不难发现: \[F(n)=\sum_{n|d}f(d)=\lfloo…