给定两个数b,d,问[1,b]和[1,d]区间上有多少对互质的数.(x,y)和(y,x)算一个. 对于[1,b]部分,用欧拉函数直接求.对于大于b的部分,求n在[1,b]上有多少个互质的数,用容斥原理. 主要学习容斥原理的写法,本题使用DFS.容斥原理复杂度比较高,是指数复杂度. 输出长整型不能用lld,必须用I64d. #include<stdio.h> #include<iostream> #include<stdlib.h> #include<string.…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5106    Accepted Submission(s): 1833 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): 9811    Accepted Submission(s): 3682 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 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…

F - GCD Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1695 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 c…

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 number of diffe…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1695 题目解析: 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. 题目又说a==c==1,所以就是求[1,b]与[1,d]中gcd等于k的个数,因为若gcd(x,y)==z,那么gcd(x/z,y/z)==1,又因为不是z的倍数的肯定不是,所以不是z的倍数的可以直接去…

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…

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…

问题:求1~r中有多少个数与n互素. 对于这个问题由容斥原理,我们有3种写法,其实效率差不多.分别是:dfs,队列数组,位运算. 先说说位运算吧: 用二进制1,0来表示第几个素因子是否被用到,如m=3,三个因子是2,3,5,则i=3时二进制是011,表示第2.3个因子被用到 LL Solve(LL n,LL r) { vector<LL> p; for(LL i=2; i*i<=n; i++) { if(n%i==0) { p.push_back(i); while(n%i==0) n/…