Spoj-BITDIFF Bit Difference
Given an integer array of N integers, find the sum of bit differences in all the pairs that can be formed from array elements. Bit difference of a pair (x, y) is the count of different bits at the same positions in binary representations of x and y. For example, bit difference for 2 and 7 is 2. Binary representation of 2 is 010 and 7 is 111 (first and last bits differ in two numbers).
Input
Input begins with a line containing an integer T(1<=T<=100), denoting the number of test cases. Then T test cases follow. Each test case begins with a line containing an integer N(1<=N<=10000), denoting the number of integers in the array, followed by a line containing N space separated 32-bit integers.
Output
For each test case, output a single line in the format Case X: Y, where X denotes the test case number and Y denotes the sum of bit differences in all the pairs that can be formed from array elements modulo 10000007.
Example
Input:
1
4
3 2 1 4 Output:
Case 1: 22
求和:对任意一对数(a,b)二进制展开,这两个数有多少位是01不同的
把每一位拆开来,假设第i位是1的有bit[i]个,答案就是Σ2*bit[i]*(n-bit[i])
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
#define mod 10000007
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n;
LL ans;
int a[];
int bit[];
inline void work(int cur)
{
n=read();
for (int i=;i<=n;i++)a[i]=read();
ans=;
memset(bit,,sizeof(bit));
for (int i=;i<=n;i++)
for (int j=;j<=;j++)
if (a[i] & (<<j))bit[j]++;
for (int j=;j<=;j++)
ans+=(LL)bit[j]*(n-bit[j])*;
printf("Case %d: %lld\n",cur,ans%mod);
}
int main()
{
int T=read(),tt=;
while (T--)work(++tt);
}
Spoj BITDIFF
Spoj-BITDIFF Bit Difference的更多相关文章
- SPOJ - BITDIFF: Bit Difference [神妙の预处理]
tags:[数学][预处理]题解:我们用一种巧妙的预处理姿势:记录下每一个数位上分别出现了多少个1.如果第i个数位上出现了cnt[i]个1,那么,在这个数位上产生的"差异值"为:2 ...
- POJ 1511 Invitation Cards / UVA 721 Invitation Cards / SPOJ Invitation / UVAlive Invitation Cards / SCU 1132 Invitation Cards / ZOJ 2008 Invitation Cards / HDU 1535 (图论,最短路径)
POJ 1511 Invitation Cards / UVA 721 Invitation Cards / SPOJ Invitation / UVAlive Invitation Cards / ...
- Java 堆内存与栈内存异同(Java Heap Memory vs Stack Memory Difference)
--reference Java Heap Memory vs Stack Memory Difference 在数据结构中,堆和栈可以说是两种最基础的数据结构,而Java中的栈内存空间和堆内存空间有 ...
- BZOJ 2588: Spoj 10628. Count on a tree [树上主席树]
2588: Spoj 10628. Count on a tree Time Limit: 12 Sec Memory Limit: 128 MBSubmit: 5217 Solved: 1233 ...
- SPOJ DQUERY D-query(主席树)
题目 Source http://www.spoj.com/problems/DQUERY/en/ Description Given a sequence of n numbers a1, a2, ...
- What's the difference between a stub and mock?
I believe the biggest distinction is that a stub you have already written with predetermined behavio ...
- SPOJ GSS3 Can you answer these queries III[线段树]
SPOJ - GSS3 Can you answer these queries III Description You are given a sequence A of N (N <= 50 ...
- 【填坑向】spoj COT/bzoj2588 Count on a tree
这题是学主席树的时候就想写的,,, 但是当时没写(懒) 现在来填坑 = =日常调半天lca(考虑以后背板) 主席树还是蛮好写的,但是代码出现重复,不太好,导致调试的时候心里没底(虽然事实证明主席树部分 ...
- [转载]Difference between <context:annotation-config> vs <context:component-scan>
在国外看到详细的说明一篇,非常浅显透彻.转给国内的筒子们:-) 原文标题: Spring中的<context:annotation-config>与<context:componen ...
随机推荐
- LeetCode Add and Search Word - Data structure design (trie树)
题意:实现添加单词和查找单词的作用,即实现字典功能. 思路:'.' 可以代表一个任何小写字母,可能是".abc"或者"a.bc"或者"abc.&quo ...
- NBUT 1116 Flandre's Passageway (LIS变形)
题意: 给一个有n*m格子的矩形,设每格边长100,要从(1,1)走到(n,m)需要耗(n+m)*100,但是其中有一些格子是可以直接穿过的,也就是走对角线,是100*根号2长,给出k个可以穿过的格子 ...
- 洛谷 P2936 [USACO09JAN]全流Total Flow
题目描述 Farmer John always wants his cows to have enough water and thus has made a map of the N (1 < ...
- Caused by: java.lang.ClassNotFoundException: org.springframework.boot.system.JavaVersion
Caused by: java.lang.ClassNotFoundException: org.springframework.boot.system.JavaVersion Invalid pro ...
- python小括号( )与中括号 [ ]
在python中小括号()表示的是tuple元组数据类型,元组是一种不可变序列. >>> a = (1,2,3) >>> a (1, 2, 3) >>& ...
- VC-基础:隐藏不安全函数的warning-_CRT_SECURE_NO_WARNINGS
>tmp.cpp(): warning C4996: 'strcpy': This function or variable may be unsafe. Consider using strc ...
- windows搭建gcc开发环境(msys2) objdump
前言 可能你并不太了解msys2,但是作为一个程序员,你一定知道mingw,而msys2就集成了mingw,同时msys2还有一些其他的特性,例如包管理器等. msys2可以在windows下搭建一个 ...
- linux or msys2设置网络代理
在文件 .bashrc 中添加 export http_proxy="proxy IP:port" 如 export http_proxy="192.168.0.1:80 ...
- hibernate3缓存(hibernate)
一级缓存:当应用程序调用Session 的save() .update() .savaeOrUpdate() .get() 或load() ,以及调用查询接口的list() .iterate() 或f ...
- word2vec 中的数学原理详解(二)预备知识
版权声明:本文为博主原创文章,未经博主允许不得转载. https://blog.csdn.net/peghoty/article/details/37969635 https://blog.csdn. ...