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…
[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…
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: 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…
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…
A. Boy or Girl 模拟题意. B. Easy Number Challenge 筛素数,预处理出\(d_i\). 三重循环枚举. C. LCM Challenge 打表找规律. 若\(n\)为奇数,则答案为\((n-2)(n-1)n\). 若\(n\)为偶数,则答案为\(max((n-3)(n-2)(n-1),(n-2)(n-1)n)\) D. Let's Play Osu! \[\prod_{i=l}^{r}{p_i}\cdot (r-l+1)^2\]相当于\([l,r]\)之间任…
