Gym - 101911C Bacteria (规律题)
传送门:点我
Time limit2000 ms
Memory limit262144 kB
Recently Monocarp has created his own mini-laboratory!
The laboratory contains nn bacteria. Monocarp knows that he can merge any two bacteria having equal sizes, and the resulting bacterium will have the size equal to the sum of sizes of merged bacteria. For example, if two bacteria having sizes equal to 77 merge, one bacterium with size 1414 is the result.
It becomes hard to watch for many bacteria, so Monocarp wants to merge all of them into one bacterium. It may not be possible to do this with the bacteria Monocarp has, so he can buy any number of bacteria of any possible integer sizes in a special store.
You have to determine the minimum number of bacteria Monocarp has to buy to merge them with the nn bacteria his laboratory contains into exactly one bacterium.
Input
The first line contains one integer nn (1≤n≤2⋅105)(1≤n≤2⋅105) — the number of bacteria Monocarp's laboratory contains.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤109)(1≤ai≤109), where aiai is the size of the ii-th bacterium in the laboratory.
Output
If it is impossible to merge the bacteria (possibly after buying some) into only one bacterium, print -1.
Otherwise print the minimum number of bacteria Monocarp has to buy to merge them with the nn bacteria his laboratory contains into exactly one bacterium.
Examples
2
1 4
2
3
3 6 9
-1
7
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1
Note
In the first example Monocarp should buy one bacterium having size 11 and one bacterium having size 22. Then Monocarp will have 44 bacteria having sizes [1,4,1,2][1,4,1,2]. Then two bacteria having sizes 11 can be merged into one having size 22. Then Monocarp will have 33 bacteria having sizes [2,4,2][2,4,2]. Then two bacteria having sizes 22 can be merged into one having size 44. Then Monocarp will have 22 bacteria having sizes [4,4][4,4], which can be merged into one having size 88.
In the second example no matter which bacteria Monocarp will buy, he cannot merge all his bacteria.
In the third example Monocarp needs to buy one bacterium having size 1000000000.
题意:给N个数,只有相同的两个数才能结合起来,加在一起。比如说2只有和2结合,成为4。
比如样例1的 1 4 ,我们需要加一个1,和原来的1结合,成为2,然后因为2没有可以组合的,所以再加一个2,组合成4,4和原来的4组合成为8。
询问的就是能不能通过添加一些数,最后能组合成1个数。如果不能输出-1,如果能输出还缺少的数的个数。
思路:
简单推理一下会发现,从小到大取,每2个数,后一个如果不是前一个的倍数,肯定不行。后一个除以前一个不是2次幂,也不行。
然后就是简单的推上去合并的问题了。这个过程用队列模拟一下。
代码:
#include<bits/stdc++.h>
using namespace std;
#define LL long long
using namespace std;
int main()
{
priority_queue<LL,vector<LL>,greater<LL> >q;
int n;
scanf("%d",&n);
for(int i = ; i < n ; i++){
LL x;
scanf("%lld",&x);
q.push(x);
}
LL ans = ;
while(q.size()>=){
LL f1 = q.top();q.pop();
LL f2 = q.top();q.pop();
if(f2%f1 != )return puts("-1"),;
LL d = f2/f1;
if((d&(d-))!=) return puts("-1"),;
q.push(f2*2LL);
ans += (LL)log2(f2*1.0/f1);
}
printf("%lld\n",ans);
}
/*
7
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 3
3 6 9 2
1 4
*/
Gym - 101911C Bacteria (规律题)的更多相关文章
- Bacteria (Gym - 101911C)
2018-2019 ACM-ICPC, NEERC, Southern Subregional Contest, Qualification Stage Bacteria Gym - 101911C ...
- LightOJ1010---Knights in Chessboard (规律题)
Given an m x n chessboard where you want to place chess knights. You have to find the number of maxi ...
- Codeforces - 规律题 [占坑]
发现自己容易被卡水题,需要强行苟一下规律题 CF上并没有对应的tag,所以本题集大部分对应百毒搜索按顺序刷 本题集侧重于找规律的过程(不然做这些垃圾题有什么用) Codeforces - 1008C ...
- http://codeforces.com/gym/100623/attachments E题
http://codeforces.com/gym/100623/attachments E题第一个优化它虽然是镜像对称,但它毕竟是一一对称的,所以可以匹配串和模式串都从头到尾颠倒一下第二个优化,与次 ...
- ACM_送气球(规律题)
送气球 Time Limit: 2000/1000ms (Java/Others) Problem Description: 为了奖励近段时间辛苦刷题的ACMer,会长决定给正在机房刷题的他们送气球. ...
- hdoj--1005--Number Sequence(规律题)
Number Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- http://codeforces.com/gym/100623/attachments H题
http://codeforces.com/gym/100623/attachments H题已经给出来的,包括后来添加的,都累加得到ans,那么从1-ans都是可以凑出来的,如果ans<a[n ...
- Codeforces Gym 100531G Grave 水题
Problem G. Grave 题目连接: http://codeforces.com/gym/100531/attachments Description Gerard develops a Ha ...
- 洛谷 P1876 开灯(思维,枚举,规律题)
P1876 开灯 题目背景 该题的题目是不是感到很眼熟呢? 事实上,如果你懂的方法,该题的代码简直不能再短. 但是如果你不懂得呢?那...(自己去想) 题目描述 首先所有的灯都是关的(注意是关!),编 ...
随机推荐
- 白鹭引擎 - 遮罩( Rectangle )
1: 矩形遮罩 class Main extends egret.DisplayObjectContainer { /** * Main 类构造器, 初始化的时候自动执行, ( 子类的构造函数必须调用 ...
- 图文详解AO打印(端桥模式)(转)
一.概述 AO打印是英文Active-Online Print的简称,也称主动在线打印.打印前支持AO通讯协议的AO打印机首先通过普通网络与C-Lodop服务保持在线链接,网页程序利用JavaSc ...
- IIS asp 401.1错误
asp程序使用非匿名帐户运行时因用户名前带了计算机名会导致出现401.1错误,只要直接输入用户名即可,不要带计算机名.
- WPF 透明窗体
窗体属性中设置:Background="Transparent" AllowsTransparency="True" WindowStyle="Non ...
- dshow采集过程
捕捉静态图片常用的filter是Sample Graber filter,它的用法参考手册.然后将捕捉filter的静态PIN连接到Sample Grabber,再将Sample Grabber连接到 ...
- html 基础之canvas 和 localStorage
1,建立一个canvas 画布: <!DOCTYPE html> <html lang="en"> <head> <meta charse ...
- jQuery+php实现二级联动
public function liandong(){ $arr = Db::table("city")->where("pid=0")->sele ...
- 开发一个FTP软件
一.开发一个多并发的FTP server 需求: .允许同时支持多用户在线 .用户认证 .用户空间配额 .权限限制 .可上传下载.上传下载过程中显示进度条 .用户可远程切换目录.查看服务端文件列表等 ...
- ArcGIS案例学习笔记-CAD数据自动拓扑检查
ArcGIS案例学习笔记-CAD数据自动拓扑检查 联系方式:谢老师,135-4855-4328,xiexiaokui#qq.com 功能:针对CAD数据,自动进行拓扑检查 优点:类别:地理建模项目实例 ...
- http://wenku.baidu.com/view/26afdb8371fe910ef12df8ccRevit采用DWG和FBX两种格式导入3D max方法的总结
2.DWG 属性——导出为 ACIS 实体(如果选择导出为多边形网格,则将每个图元导出为 由多个多边形组成的对象,这些多边形相互连接或组成 “ 网格 ” .在导出到 DWG 文件以用 于 Ma ...