/* HDU5976 Detachment http://acm.hdu.edu.cn/showproblem.php?pid=5976 数论 等差数列 * * */ #include <cstdio> #include <algorithm> #include <vector> using namespace std; ; ; vector<long long> v; long long f[Nmax]; long long g[Nmax]; long l…

Detachment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 392    Accepted Submission(s): 131 Problem Description In a highly developed alien society, the habitats are almost infinite dimensiona…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5976 Detachment Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 In a highly developed alien society, the habitats are almost infinite dimensional space. In the histo…

hdu 5976 Detachment题目连接 题意: 给定一个自然数x,让你给出一种拆分方式n=a1+a2+...(ai≠aj),使得每个小部分的乘积s=a1*a2*...最大 解题思路: 我们要乘积最大,那么我们把n尽可能的拆分,如果题目不要求ai≠aj,那么我们将x拆分成什么最好呢?显然拆成2和3(2为偶数,3为奇数,2x+3y可以表示大于1的所有正整数)是最好的,顺便提下3最多的时候最优,为什么不是2,将,6拆成3个2乘积为8,而将6拆成3乘积为9,(不会证) 那这样我们就把n分成x个2…

HDU 5976 Detachment(拆分) 00 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)   Problem Description - 题目描述 In a highly developed alien society, the habitats are almost infinite dimensional space. In the history of this planet,there is an o…

F - Detachment In a highly developed alien society, the habitats are almost infinite dimensional space. In the history of this planet,there is an old puzzle. You have a line segment with x units’ length representing one dimension.The line segment can…

Detachment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 570    Accepted Submission(s): 192 Problem Description In a highly developed alien society, the habitats are almost infinite dimensiona…

题意:输入一个x,将x拆分成一些小的数(这些数不能相同,即x=a1+a2+......   ai!=aj when i!=j),然后这些数相乘得到一个成积(s=a1*a2*......),求最大的乘积s: 思路:考虑最简单的做法便是贪心,很明显将一个数分的越小,这个乘积越大,那么对于给的x 先找2+3+4+....+n<=x 找到最大的n  如果和小于x ,那么将n右移一个数(n->n+1)  如果和还小于x继续将n-1右移......知道和x相等时,输出s   这样做时间复杂度很高,那么得优…

题意:给定n,要求构造若干个各不相同且和为n的正整数使得它们的乘积最大 T<=1e6,1<=n<=1e9 思路:From https://blog.csdn.net/qq_34374664/article/details/53466435 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> typede…

题目链接 题意 将\(x\)拆成\(a_1+a_2+...+\)的形式,且\(a_1\lt a_2\lt...\),使得\(a_1*a_2*...\)取到最大值 思路 大胆猜结论. 首先拆分的形式中肯定不能有\(1\). 于是预处理出前缀和\(a[i]=\sum_{k=2}^{i}k\), 找到\(\geq x\)的最小的\(a[id]\),接下来: 如果\(a[id]==x\),意味着\(2+3+...+id=x\),那么答案就是\(2*3*...*id=factorial(id)\) 否则,…