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
- Scrapy-01-追踪爬取
目的:利用scrapy完成盗墓笔记小说的抓取 创建项目: scrapy startproject books cd books scrapy genspider dmbj 编写p ...
- 【python】利用jieba中文分词进行词频统计
以下代码对鲁迅的<祝福>进行了词频统计: import io import jieba txt = io.open("zhufu.txt", "r" ...
- Java研发技术学习路线
Java研发技术成长路线 作为一名Java研发者,深感Java技术的学习是一个漫长过程,从一名Java菜鸟开始,加之持之以恒的耐心和脚踏实地的精神,不间断理论的学习,不停止技术实践,终成为一名技术佼佼 ...
- java day04记录
本文主要记录arr数组用法.count计算.arr倒排序技巧案例 package day4homework; import java.util.Scanner; /* 从键盘上输入10个整数,合法值位 ...
- tf.nn.dropout 激活函数
tf.nn.dropout(x,keep_prob,noise_shape=None,seed=None,name=None) 参数: x:一个浮点型Tensor. keep_prob:一个标量Ten ...
- nghttp2 交叉编译
touch run.sh chmod 755 run.sh mkdir build cd build ../run.sh run.sh #!/bin/bash #cd build ../configu ...
- push和appendChild的区别
概述:绑定事件(push和appendChild用法相似:但是一个是控制数组,一个是控制元素节点)用法:1.数组1的更改后的长度 = 数组1.push();//用来控制数组,在数组最后面插入项,返回数 ...
- HashMap主要方法源码分析(JDK1.8)
本篇从HashMap的put.get.remove方法入手,分析源码流程 (不涉及红黑树的具体算法) jkd1.8中HashMap的结构为数组.链表.红黑树的形式 (未转化红黑树时) (转 ...
- tensorflow--filter、strides
最近还在看<TensorFlow 实战Google深度学习框架第二版>这本书,根据第六章里面对于卷基层和池化层的介绍可以发现,在执行 tf.nn.conv2d 和 tf.nn.max_po ...
- OkHttp 优雅封装 HttpUtils 之 气海雪山初探
曾经在代码里放荡不羁,如今在博文中日夜兼行,只为今天与你分享成果.如果觉得本文有用,记得关注我,我将带给你更多. 介绍 HttpUtils 是近期开源的对 OkHttp 轻量封装的框架,它独创的异步预 ...