Mophues 题意:给出n, m, p,求有多少对a, b满足gcd(a, b)的素因子个数<=p,(其中1<=a<=n, 1<=b<=m) 有Q组数据:(n, m, P <= 5×105. Q <=5000). 参考:ACdreamers 思路:对于hdu1695 GCD来说,由于只需要求gcd = k的个数,所以我们可以按照不优化莫比乌斯公式直接求,这样求解的时间复杂度为O(n); 若这题我们也不优化,答案累加需要双重循环: rep1(i,,n){ if(n…

Mophues Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327670/327670 K (Java/Others) Total Submission(s): 980    Accepted Submission(s): 376 Problem Description As we know, any positive integer C ( C >= 2 ) can be written as the multiply of…

分析: http://blog.csdn.net/acdreamers/article/details/12871643 分析参见这一篇 http://wenku.baidu.com/view/fbe263d384254b35eefd34eb.html 分块看这一篇 #include<cstdio> #include<cstring> #include<queue> #include<cstdlib> #include<algorithm> #i…

Mophues \[ Time Limit: 10000 ms\quad Memory Limit: 262144 kB \] 题意 求出满足 \(gcd\left(a,b\right) = k\),其中\(1\leq a\leq n,1\leq b \leq m\)且 \(k\) 的因子数 \(\leq P\) 思路 \(g\left(x\right)\) 表示 \(gcd\left(a, b\right) | x\) 的对数 \(f\left(x\right)\) 表示 \(gcd\left…

链接 求[1,n] 和 [1,m]中有多少对数的GCD的素因子个数小于等于p 直接暴力做特定超时,所以我们想办法预处理,对于p大于18(1到5e5的最大素数因子个数)的情况,每一对都满足条件,O(1)得结果. p<=18时,预处理sum[i][j]sum[i][j]表示所有能整除i的质因子个数<=j的(x,y)的对数,然后求一个前缀和,ans=∑⌊n/d⌋∗⌊m/d⌋∗∑k|dμ(d/k) //#pragma comment(linker, "/stack:200000000&quo…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4746 题意: 1≤x,y≤n , 求gcd(x,y)分解后质因数个数小于等k的(x,y)的对数. 分析: 莫比乌斯反演. 还是一个套路,我们设 f(d):满足gcd(x,y)=d且x,y均在给定范围内的(x,y)的对数. F(d):满足d|gcd(x,y)且x,y均在给定范围内的(x,y)的对数. 显然F(x)=[n/x]∗[m/x],反演后我们得到 f(x)=∑x|dμ(d/x)[n/d]∗[m…

题意: \(Q\leq5000\)次询问,每次问你有多少对\((x,y)\)满足\(x\in[1,n],y\in[1,m]\)且\(gcd(x,y)\)的质因数分解个数小于等于\(p\).\(n,m,p\leq5e5\). 思路: 题目即求 \[\sum_{k}\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=k]\quad,k满足质因数个数\leq p \] 令\(f(n)\)为\(gcd\)为\(n\)的对数,\(F(n)\)为\(gcd\)为\(n\)倍数的对数. 由莫比…

莫比乌斯反演.先初始化出所有数有多少个质因子和mobius.然后处理mob_sum[ i ][ j ],表示当公因子的因子个数小于等于 j 个的mobius前 i 项和.然后分块求和即可. 分块处理部分见(不会莫比乌斯反演的同学也可以去这里学一下)http://wenku.baidu.com/view/fbe263d384254b35eefd34eb.html. #include<algorithm> #include<iostream> #include<cstring&g…

Coprime Arrays CodeForces - 915G Let's call an array a of size n coprime iff gcd(a1, a2, ..., *a**n) = 1, where gcd* is the greatest common divisor of the arguments. You are given two numbers n and k. For each i (1 ≤ i ≤ k) you have to determine the…

Code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 300    Accepted Submission(s): 124 Problem Description WLD likes playing with codes.One day he is writing a function.Howerver,his computer b…