hdu1695 GCD2 容斥原理求x属于[1,b]与y属于[1,d]，gcd(x,y)=k的对数。(5,7)与(7,5)看作同一对。

【hdu1695 GCD2 容斥原理求x属于[1,b]与y属于[1,d]，gcd(x,y)=k的对数。(5,7)与(7,5)看作同一对。】的更多相关文章

hdu1695 GCD2 容斥原理求x属于[1,b]与y属于[1,d]，gcd(x,y)=k的对数。(5,7)与(7,5)看作同一对。

GCD Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Given integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest…

GCD 莫比乌斯反演给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的数对(x,y)有多少对.

/** 题目:GCD 链接:https://vjudge.net/contest/178455#problem/E 题意: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. 1<=N<=10^7 思路: 定义: f(n)表示gcd(x,y)==n的对数. F(n)表示n|gcd(x,y)的对数. 枚举n以内的素数p: f(p) = sigma(mu[d/p]*F(d)) (p|d) F(d) = (n/d)*(n/d); N/10*sqrt(N)复…

hdu2588 GCD 给定n，m。求x属于[1,n]。有多少个x满足gcd(x,n)>=m; 容斥或者欧拉函数

GCD Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),,)=,(,)=. (a,b) can be e…

容斥原理求M以内有多少个跟N是互质的

开始系统的学习容斥原理!通常我们求1-n中与n互质的数的个数都是用欧拉函数! 但如果n比较大或者是求1-m中与n互质的数的个数等等问题,要想时间效率高的话还是用容斥原理! 本题是求[a,b]中与n互质的数的个数,可以转换成求[1,b]中与n互质的数个数减去[1,a-1]与n互质的数的个数. #include<iostream> #include<cstdio> #include<cstring> using namespace std; #define LL lon…

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题目给定两个区间[1, b], [1, d],统计数对的个数(x, y)满足: \(x \in [1, b]\), \(y \in [1, d]\) ; \(gcd(x, y) = k\) HDU1695 题解我们观察式子\(gcd(x,y)=k\) 很显然,\(gcd(x/k, y/k) = 1\) 我们令b < d,令x<y(避免重复计数) 分类讨论. 1) y < b 可以看出答案就是\(\sum_{i \in [1, b]} \phi(i)\) 2)\(y \in [b, d…

ACM学习历程—HDU1695 GCD（容斥原理 || 莫比乌斯）

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) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total num…