codeforce 272B Dima and Sequence
B. Dima and Sequence
Dima got into number sequences. Now he's got sequence a1, a2, ..., an, consisting of n positive integers. Also, Dima has got a function f(x), which can be defined with the following recurrence:
- f(0) = 0;
- f(2·x) = f(x);
- f(2·x + 1) = f(x) + 1.
Dima wonders, how many pairs of indexes (i, j) (1 ≤ i < j ≤ n) are there, such that f(ai) = f(aj). Help him, count the number of such pairs.
Input
The first line contains integer n (1 ≤ n ≤ 105). The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).
The numbers in the lines are separated by single spaces.
Output
In a single line print the answer to the problem.
Please, don't use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64dspecifier.
Examples
input
3
1 2 4
output
3
input
3
5 3 1
output
1
Note
In the first sample any pair (i, j) will do, so the answer is 3.
In the second sample only pair (1, 2) will do.
打表100条,找到规律。
偶数找规律,计数找祖宗。
#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
//---------------------------------Sexy operation--------------------------//
#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define debug(n) printf("%d_________________________________\n",n);
#define speed ios_base::sync_with_stdio(0)
#define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
//-------------------------------Actual option------------------------------//
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Swap(a,b) a^=b^=a^=b
#define Max(a,b) (a>b?a:b)
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define mp(a,b) make_pair(a,b)
#define pb(n) push_back(n)
//--------------------------------constant----------------------------------//
#define INF 0x3f3f3f3f
#define maxn 10000000
#define esp 1e-9
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef long long ll;
//___________________________Dividing Line__________________________________//
long long a[40]= {0};
int main()
{
int n,x;
cini(n);
while(n--)
{
int sum=0;
cini(x);
while(x){
if(x&1){
sum++;
x=(x-1)/2;
}
else x/=2;
}
a[sum]++;
}
long long ans=0;
for(int i=0; i<40; i++)
ans+=a[i]*(a[i]-1)/2;
cout<<ans<<endl;
return 0;
}
codeforce 272B Dima and Sequence的更多相关文章
- Codeforce 438D-The Child and Sequence 分类: Brush Mode 2014-10-06 20:20 102人阅读 评论(0) 收藏
D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input st ...
- codeforce 272E Dima and Horses (假DFS)
E. Dima and Horses Dima came to the horse land. There are n horses living in the land. Each horse in ...
- [CodeForce]358D Dima and Hares
有N<3000只宠物要喂,每次只能喂一只,每喂一只宠物,宠物的满足度取决于: 1 紧靠的两个邻居都没喂,a[i] 2 邻居中有一个喂过了,b[i] 3 两个邻居都喂过了,c[i] 把所有宠物喂一 ...
- codeforce gym/100495/problem/K—Wolf and sheep 两圆求相交面积 与 gym/100495/problem/E—Simple sequence思路简述
之前几乎没写过什么这种几何的计算题.在众多大佬的博客下终于记起来了当时的公式.嘚赶快补计算几何和概率论的坑了... 这题的要求,在对两圆相交的板子略做修改后,很容易实现.这里直接给出代码.重点的部分有 ...
- Codeforces Round #167 (Div. 2) D. Dima and Two Sequences 排列组合
题目链接: http://codeforces.com/problemset/problem/272/D D. Dima and Two Sequences time limit per test2 ...
- Codeforce 水题报告(2)
又水了一发Codeforce ,这次继续发发题解顺便给自己PKUSC攒攒人品吧 CodeForces 438C:The Child and Polygon: 描述:给出一个多边形,求三角剖分的方案数( ...
- Two progressions CodeForce 125D 思维题
An arithmetic progression is such a non-empty sequence of numbers where the difference between any t ...
- CodeForce 577B Modulo Sum
You are given a sequence of numbers a1, a2, ..., an, and a number m. Check if it is possible to choo ...
- codeforce 1311 C. Perform the Combo 前缀和
You want to perform the combo on your opponent in one popular fighting game. The combo is the string ...
随机推荐
- npm install报错:chromedriver@2.27.2 install: node install.js
报错: 刚开始以为是 node 或 npm 版本问题,前前后后折腾了好久,终于解决了 解决: 如果执行过npm install,先删除 node_modules 文件夹,不然运行的时候可能会报错 执行 ...
- Struts2-学习笔记系列(15)-ajax支持和JSON
7.1stream类型的result 使用stream就无需jsp页面,直接在action想浏览者生成指定的响应 @Override public java.lang.String execute() ...
- HBase协处理器加载的三种方式
本文主要给大家罗列了HBase协处理器加载的三种方式:Shell加载(动态).Api加载(动态).配置文件加载(静态).其中静态加载方式需要重启HBase. 我们假设我们已经有一个现成的需要加载的协处 ...
- Redis之quicklist源码分析
一.quicklist简介 Redis列表是简单的字符串列表,按照插入顺序排序.你可以添加一个元素到列表的头部(左边)或者尾部(右边). 一个列表最多可以包含 232 - 1 个元素 (4294967 ...
- thinkphp5.0远程执行漏洞
0x01 漏洞简介 由于ThinkPHP5 框架控制器名 没有进行足够的安全监测,导致在没有开启强制路由的情况下,可以伪装特定的请求可以直接Getshell(可以控制服务器) 0x02 环境搭建 Ph ...
- 前端学习笔记 --ES6新特性
前言 这篇博客是我在b站进行学习es6课程时的笔记总结与补充. 此处贴出up主的教程视频地址:深入解读ES6系列(全18讲) 1.ES6学习之路 1.1 ES6新特性 1. 变量 2. 函数 3. 数 ...
- 浅谈Vector
浅谈Vector 在之前的文章中,我们已经说过线程不安全的ArrayList和LinkedList,今天我们来讲讲一个线程安全的列表容器,他就是Vector,他的底层和ArrayList一样使用数组来 ...
- B - How Many Equations Can You Find dfs
Now give you an string which only contains 0, 1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9.You are asked to add the sig ...
- 全网最全最细的jmeter接口测试教程以及接口测试流程详解
一.Jmeter简介 Jmeter是由Apache公司开发的一个纯Java的开源项目,即可以用于做接口测试也可以用于做性能测试. Jmeter具备高移植性,可以实现跨平台运行. Jmeter可以实 ...
- 9. 弹出键盘挡住input
1.) react 中 <input className="inp3" placeholder="密码" type="password" ...