GCD and LCM Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2024 Accepted Submission(s): 904 Problem Description Given two positive integers G and L, could you tell me how many solutions of (
J - Worker Avin meets a rich customer today. He will earn 1 million dollars if he can solve a hard problem. There are n warehouses and m workers. Any worker in the i-th warehouse can handle ai orders per day. The customer wonders whether there exists
C. Neko does Maths time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher. Neko ha
1.gcd 递归实现 int gcd(int i,int j){ if(j==0) return i; else return gcd(j,i%j);} 2.lcm int gcd(int i,int j){if(j==0) return i;else return gcd(j,i%j);} int lcm(int i ,int j) { return i*j/gcd(i*j); } 3. E - Wolf and Rabbit 兔子坑问题 There is a hill with n hole
II U C ONLINE C ON TEST Problem D: GCD LCM Input: standard input Output: standard output The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smalle
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3071 题目大意: 给定一个长度为n的序列m次操作,操作的种类一共有三种 查询 L :查询一个区间的所有的数的最小公倍数modp G :查询一个区间的所有的数的最大公约数modp 修改 C :将给定位置的值修改成x 解题思路: 注意数据范围,每个数字不超过100,所以100以内的质因子最多25个,如果直接求解lcm和gcd的话,long long也是存不下的,所以采用存储质因子的指数,但是如果每个节