GCD 题意:输入5个数a,b,c,d,k;(a = c = 1, 0 < b,d,k <= 100000);问有多少对a <= p <= b, c <= q <= d使得gcd(p,q) = k; 注:对于(p,q)和(q,p)只算一次: 思路:由于遍历朴素求两个数的gcd的时间复杂度为O(n^2*log(n)),朴素算法遍历搜索在判断累加,所以效率很低: 资料   NanoApe's Blog   ACdreamers 莫比乌斯反演:利用整与分之间的可逆来由整体利用…

分析:简单的莫比乌斯反演 f[i]为k=i时的答案数 然后就很简单了 #include<iostream> #include<algorithm> #include<set> #include<vector> #include<queue> #include<cstdlib> #include<cstdio> #include<cstring> #include<cmath> using names…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9765    Accepted Submission(s): 3652 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

[BZOJ 2820] YY的gcd(莫比乌斯反演+数论分块) 题面 给定N, M,求\(1\leq x\leq N, 1\leq y\leq M\)且gcd(x, y)为质数的(x, y)有多少对.q组询问 分析 我们要求的是 \[\sum_{p \in P} \sum_{i=1}^n \sum_{j=1}^m [gcd(i,j)=p]\](大写P表示质数集合) 根据\(kgcd(i,j)=gcd(ki,kj)\), \[原式=\sum_{p \in P} \sum_{i=1}^{\lfloo…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 17212    Accepted Submission(s): 6637 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x,…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4141    Accepted Submission(s): 1441 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4291    Accepted Submission(s): 1502 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

题目链接 这题求[1,n],[1,m]gcd为k的对数.而且没有顺序. 设F(n)为公约数为n的组数个数 f(n)为最大公约数为n的组数个数 然后在纸上手动验一下F(n)和f(n)的关系,直接套公式就好了.注意要删去重复的. 关于 莫比乌斯反演 的结论 ACdreamers大神的相关博客 莫比乌斯反演  莫比乌斯反演与最大公约数 #include<bits/stdc++.h> using namespace std; typedef long long LL; const int maxn=1…

题链: http://acm.hdu.edu.cn/showproblem.php?pid=1695 题解: 容斥. 莫比乌斯反演,入门题. 问题化简:求满足x∈(1~n)和y∈(1~m),且gcd(x,y)=1的(x,y)的对数. 下文默认$n \leq m$ 1.容斥 (先写了一个的裸的容斥.) 令$f(k)为gcd(x,y)=\lambda k的(x,y)的对数$ $ANS=f(0种质数的积)-f(1种质数的积)+f(2种质数的积)-\cdots+(-1)^mf(m种质数的积)$ 代码:…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1695 题意: 对于 a, b, c, d, k . 有 x 属于 [a, b],  y 属于 [c, d], 求 gcd(x, y) = k 的 x, y 的对数 . 其中 a = b = 1 . 注意: (x, y), (y, x) 算一种情况 . 思路: 莫比乌斯反演 可以参考一下: http://blog.csdn.net/lixuepeng_001/article/details/5057…