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
- 10.1 io流--ASCII码表
day2.8中提到 /* * +: * 做加法运算 * * 字符参与加法运算,其实是拿字符在计算机中存储的数据值来参与运算的 * 'A' 65(B 66...) * 'a' 97(b 98...) * ...
- 9.2ArrayList 集合 案例,学生管理系统
循环的使用 添加学生:while嵌套for,for设置变量,内嵌if更新变量.if语句判断变量值 修改学生:for循环内嵌if,获取循环中的某个值. package day9_ArrayList.AL ...
- cxx signal信号捕获
kill -9 [pid] 该信号不能被捕获 #include <iostream> #include <csignal> static void vSignalHandler ...
- ClickOnce的安装路径
win 7下C:\Users\Administrator.U5G4L4PUY34SH5C\AppData\Local\Apps\2.0\KPVZOAYK.0JE\56B55RCH.A7A\winr.. ...
- Hadoop学习笔记(1)-Hadoop在Ubuntu的安装和使用
由于小编在本学期有一门课程需要学习hadoop,需要在ubuntu的linux系统下搭建Hadoop环境,在这个过程中遇到一些问题,写下这篇博客来记录这个过程,并把分享给大家. Hadoop的安装方式 ...
- django.template.exceptions.TemplateDoesNotExist: login.html报错
前言 在某一次按以前的步骤使用Django “django.template.exceptions.TemplateDoesNotExist: login.html”错误,在以为是html文件出 ...
- 【WPF学习】第六十七章 创建自定义面板
前面两个章节分别介绍了两个自定义控件:自定义的ColorPicker和FlipPanel控件.接下来介绍派生自定义面板以及构建自定义绘图控件. 创建自定义面板是一种特殊但较常见的自定义控件开发子集.前 ...
- 重启mysql服务
重启mysql 启动mysql: 方式一:sudo /etc/init.d/mysql start 方式二:sudo service mysql start 停止mysql: 方式一:sudo /et ...
- Android App安全渗透测试(一)
一. 实验环境搭建 1. 安装JDK 2. 安装Android Studio 3. 模拟器或真机 我的是夜神模拟器和nexus 工具 Apktool ...
- 解决cvc-complex-type.2.4.a: Invalid content was found starting with element
今天用myeclipse导入 一个项目出现后出现cvc-complex-type.2.4.a: Invalid content was found starting with element 'inf ...