CSES1081:Common Divisors 传送门 题意:找到两个gcd最大的数。n≤2e5,ai≤1e6。 一种方法是枚举i:1∼n,O(ai)把ai因数的出现次数加一。 然后i:1000000∼1,如果cnt[i]>1,输出i结束。 复杂度O(nV),2e8,可惜 CSES 的机子跑不过。 枚举倍数。
Code README This was my practice run for ICPC. There are more than 250 accepted solutions listed here. I have also added tags for some of the problems. At first try to come up with a solution by yourself. If you can't then read some article on the associated tags and try again. ...
0Tags Code README CSES Solutions Over 280 accepted solutions to theCSES Problem Set, written in C++ by Jonathan Uy(nulltype). As of December 23th, the following number of solutions have been completed: Problem TypeNumber Solved Introductory Problems19/19 ...
common_divisors.cpp counting_divisors.cpp divisor_analysis.cpp exponentation.cpp exponentation_ii.cpp sum_of_divisors.cpp Range_Queries Sorting_and_Searching Tree_Algorithms Codeforces DMOJ FacebookHackercup GoogleCodeJam GoogleKickstart IOI IZhO JOI Kattis LeetCode USACOcontests USACOtraining docs helper...
evenvalue/csesPublic NotificationsYou must be signed in to change notification settings Fork0 Star4 Folders and files Name Last commit message Last commit date Latest commit History 170 Commits .idea .clang-format .gitignore 1068.cpp 1069.cpp ...