[codeforces 235]A. LCM Challenge 试题描述 Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't want to use many numbers, so I'll choose three…

Discription Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum: Find the sum modulo 1073741824 (230). Input The first line contains three space-sep…

A. LCM Challenge 题目连接: http://www.codeforces.com/contest/235/problem/A Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't wa…

 LCM Challenge Time Limit:1000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu Submit Status Practice ACdream 1077 Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I…

LCM Challenge Time Limit: 2000/1000MS (Java/Others)    Memory Limit: 128000/64000KB (Java/Others) Submit Status Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want…

A - LCM Challenge Time Limit: 2000/1000MS (Java/Others)    Memory Limit: 128000/64000KB (Java/Others) Submit Status Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I w…

http://codeforces.com/problemset/problem/235/E (题目链接) 题意 给出${a,b,c}$,求${\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cd(ijk)}$ extra 有这样一个公式,就是约数个数和那道题的推广吧.$${\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cd(ijk)=\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^c[gcd(i,j)=gcd(i,k)=gcd…

题目:http://codeforces.com/contest/401/problem/C 题意:n个0,m个1,求没有00或111的情况. 这么简单的题..... 做题的时候脑残了...,今天,贴一下ac的代码,警示一下自己 #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespa…

Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't want to use many numbers, so I'll choose three positive integers…

找到3个不超过n的正整数(可以相同),使得它们的lcm(最小公倍数)最大. Solution 可以做得很优雅吧,但我喜欢(只会)暴力一点 根据质数密度分布性质,最后所取的这三个数一定不会比 \(n\) 小太多(实在不行就取三个质数呀),所以我们钦定一个界,然后暴力枚举取最优即可 #include <bits/stdc++.h> using namespace std; #define int long long int n; signed main() { cin>>n; int…