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 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 be split into a number of small line segments: a1,a2, … (x= a1+a2+…) assigned to different dimensions. And then, the multidimensional space has been established. Now there are two requirements for this space: 
1.Two different small line segments cannot be equal ( ai≠aj when i≠j).
2.Make this multidimensional space size s as large as possible (s= a1∗a2*...).Note that it allows to keep one dimension.That's to say, the number of ai can be only one.
Now can you solve this question and find the maximum size of the space?(For the final number is too large,your answer will be modulo 10^9+7)
 
Input
The first line is an integer T,meaning the number of test cases.
Then T lines follow. Each line contains one integer x.
1≤T≤10^6, 1≤x≤10^9
 
Output
Maximum s you can get modulo 10^9+7. Note that we wants to be greatest product before modulo 10^9+7.
 
Sample Input
1
4
 
Sample Output
4
 
Source
 
Recommend
wange2014   |   We have carefully selected several similar problems for you:  6010 6009 6008 6007 6006 
 

Statistic | Submit | Discuss | Note

题目链接:

  http://acm.hdu.edu.cn/showproblem.php?pid=5976

题目大意:

  给一个数N(N<=109),让你把它拆成若干各不相同的数Ai,ΣAi=N,要求ΠAi(累乘)最大。

题目思路:

  【贪心】

  首先肯定要把位数拆的尽量多,手写了20以内的拆法。

  发现以2为首相的递增序列累乘最大,所以我的想法就是把N拆成2+3+...+x<=n,

  先找到x,之后算一下n还多了多少,就把后面依次+1,变成2+3+...+y+(y+2)+(y+3)+...+(x+1)。

  这时候它们的累乘是最大的。

  (特殊情况是从2到x都加1之后还剩余1,这时候把最后一项再加1,变成3+4+...+x+(x+2)

 //
//by coolxxx
/*
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
//#include<stdbool.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
*/
#include<bits/stdc++.h>
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-8)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#define N 45004
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
int n,m,lll,ans;
LL a[N],ni[N];
LL mi(LL x,LL y)
{
LL z=;
while(y)
{
if(y&)z=(z*x)%mod;
x=(x*x)%mod;
y>>=;
}
return z;
}
void init()
{
int i;
a[]=;
ni[]=;
for(i=;i<N;i++)
{
a[i]=(a[i-]*i)%mod;
ni[i]=(-(mod/i)*a[mod%i])%mod;
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("1.txt","r",stdin);
// freopen("2.txt","w",stdout);
#endif
int i,j,k;
int x,y,z;
init();
for(scanf("%d",&cass);cass;cass--)
// for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
// while(~scanf("%s",s))
// while(~scanf("%d%d",&n,&m))
{
scanf("%d",&n);
if(n<)
{
printf("%d\n",n);
continue;
}
m=n+n+;
LL l,r,mid;
l=;r=;
while(l<r)
{
mid=(l+r+)>>;
if(mid*mid+mid<=m)l=mid;
else r=mid-;
}
m-=l*l+l;
m/=;
if(m==l)
{
aans=a[l]*(l+)%mod*mi(,mod-)%mod;
}
else
{
x=l+-m;
aans=a[l+]*mi(x,mod-)%mod;
}
printf("%lld\n",aans);
}
return ;
}
/*
// //
*/

HDU 5976 Detachment 【贪心】 (2016ACM/ICPC亚洲区大连站)的更多相关文章

  1. HDU 5979 Convex【计算几何】 (2016ACM/ICPC亚洲区大连站)

    Convex Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Subm ...

  2. 2016ACM/ICPC亚洲区大连站现场赛题解报告(转)

    http://blog.csdn.net/queuelovestack/article/details/53055418 下午重现了一下大连赛区的比赛,感觉有点神奇,重现时居然改了现场赛的数据范围,原 ...

  3. 2016ACM/ICPC亚洲区大连站-重现赛

    题目链接:http://acm.hdu.edu.cn/search.php?field=problem&key=2016ACM%2FICPC%D1%C7%D6%DE%C7%F8%B4%F3%C ...

  4. 2016ACM/ICPC亚洲区大连站-重现赛(感谢大连海事大学)(7/10)

    1001题意:n个人,给m对敌对关系,X个好人,Y个坏人.现在问你是否每个人都是要么是好人,要么是坏人. 先看看与X,Y个人有联通的人是否有矛盾,没有矛盾的话咋就继续遍历那些不确定的人关系,随便取一个 ...

  5. HDU 5950 Recursive sequence 【递推+矩阵快速幂】 (2016ACM/ICPC亚洲区沈阳站)

    Recursive sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Other ...

  6. HDU 5952 Counting Cliques 【DFS+剪枝】 (2016ACM/ICPC亚洲区沈阳站)

    Counting Cliques Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) ...

  7. HDU 5948 Thickest Burger 【模拟】 (2016ACM/ICPC亚洲区沈阳站)

    Thickest Burger Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)T ...

  8. HDU 5949 Relative atomic mass 【模拟】 (2016ACM/ICPC亚洲区沈阳站)

    Relative atomic mass Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Oth ...

  9. HDU 6227.Rabbits-规律 (2017ACM/ICPC亚洲区沈阳站-重现赛(感谢东北大学))

    Rabbits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total S ...

随机推荐

  1. pytorch学习 中 torch.squeeze() 和torch.unsqueeze()的用法

    squeeze的用法主要就是对数据的维度进行压缩或者解压. 先看torch.squeeze() 这个函数主要对数据的维度进行压缩,去掉维数为1的的维度,比如是一行或者一列这种,一个一行三列(1,3)的 ...

  2. 怎样从SpringMVC返回json数据

    Srping3中配置 maven依赖pom.xml 需要jackson库的依赖 <dependency> <groupId>org.codehaus.jackson</g ...

  3. HYSBZ - 2763 飞行路线(分层图最短路线)

    题目: Alice和Bob现在要乘飞机旅行,他们选择了一家相对便宜的航空公司.该航空公司一共在n个城市设有业务,设这些城市分别标记为0到n-1,一共有m种航线,每种航线连接两个城市,并且航线有一定的价 ...

  4. idea搭建SSM的maven项目(tomcat容器)

    一.创建maven的web项目 (1)选择项目的骨架 (2)写项目的坐标 (3)maven的设置 设置maven的本地仓库,以及配置文件的位置,同时点击+号,填入archetypeCatalog和in ...

  5. Centos7配置ThinkPHP5.0完整过程(一)

    在Centos中配置PHP服务器环境,首先要安装Apache的http服务,然后安装php解析环境,最后再配置ThinkPHP5.0. 首先安装HTTP sudo yum install httpd ...

  6. 使用requests+BeaBeautiful Soup爬取妹子图图片

    1. Requests:让 HTTP 服务人类 Requests 继承了urllib2的所有特性.Requests支持HTTP连接保持和连接池,支持使用cookie保持会话,支持文件上传,支持自动确定 ...

  7. python 深浅拷贝&集合

    一.深浅拷贝 1.浅拷贝,只会拷贝第一层 s = [1, 'ss', '小可爱'] s1 = s.copy() print(s1) >>> [1, 'ss', '小可爱'] s = ...

  8. java中"=="和equals方法究竟有什么区别?

    为什么会说到这个问题呢,是因为在java中遇到这个问题太常见了,无论是在写代码时还是在面试时.下面就一起探讨一下它们之间的联系与区别吧. 首先对于这样的问题,一般是先单独把一个东西说清楚,然后再说另一 ...

  9. POJ3352-Road Construction(边连通分量)

    It's almost summer time, and that means that it's almost summer construction time! This year, the go ...

  10. MyChrome制作Chrome浏览器便携版

    Google Chrome官方离线下载地址: https://api.shuax.com/tools/getchrome MyChrome下载地址: http://code.taobao.org/p/ ...