开始做 Project Euler 的练习题.网站上总共有565题,真是个大题库啊! # Project Euler, Problem 1: Multiples of 3 and 5 # If we list all the natural numbers below 10 # that are multiples of 3 or 5, we get 3, 5, 6 and 9. # The sum of these multiples is 23. # Find the sum of all
题目链接:HDOJ - 5212 题目分析 首先的思路是,考虑每个数对最终答案的贡献. 那么我们就要求出:对于每个数,以它为 gcd 的数对有多少对. 显然,对于一个数 x ,以它为 gcd 的两个数一定都是 x 的倍数.如果 x 的倍数在数列中有 k 个,那么最多有 k^2 对数的 gcd 是 x . 同样显然的是,对于两个数,如果他们都是 x 的倍数,那么他们的 gcd 一定也是 x 的倍数. 所以,我们求出 x 的倍数在数列中有 k 个,然后就有 k^2 对数满足两个数都是 x 的倍数,这
Joseph Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 47657 Accepted: 17949 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n,
Code Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 597 Accepted Submission(s): 230 Problem Description WLD likes playing with codes.One day he is writing a function.Howerver,his computer b
A - Easy $h$-index 后缀扫一下 #include <bits/stdc++.h> using namespace std; #define ll long long #define N 200010 int n; ll arr[N]; inline int work() { ll sum = ; ; --i) { sum += arr[i]; if (sum >= i) return i; } ; } int main() { while (scanf("%d