浙江省第十二届省赛 B - Team Formation
Description
For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming a two-man team from N students of his university.
Edward knows the skill level of each student. He has found that if two students with skill level A and B form a team, the skill level of the team will be A ⊕ B, where ⊕ means bitwise exclusive or. A team will play well if and only if the skill level of the team is greater than the skill level of each team member (i.e. A ⊕ B > max{A, B}).
Edward wants to form a team that will play well in the contest. Please tell him the possible number of such teams. Two teams are considered different if there is at least one different team member.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains an integer N (2 <= N <= 100000), which indicates the number of student. The next line contains N positive integers separated by spaces. The ith integer denotes the skill level of ith student. Every integer will not exceed 109.
Output
For each case, print the answer in one line.
Sample Input
2 3 1 2 3 5 1 2 3 4 5
Sample Output
1 6
题意:
输入n个数:任意两个数进行异或,问异或值大于这两个数的最大值的情况共有几种
题解:
首先将每个数的最高位哈希,然后遍历每个数。每个数转化为二进制数,二进制中只要当前位为0,那么异或此位为最高位的数一定比这两个数大。
代码:
#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <cstdlib>
#include <string>
using namespace std;
#define is_lower(c) (c>='a' && c<='z')
#define is_upper(c) (c>='A' && c<='Z')
#define is_alpha(c) (is_lower(c) || is_upper(c))
#define is_digit(c) (c>='0' && c<='9')
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define IO ios::sync_with_stdio(0);\
cin.tie();\
cout.tie();
#define For(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const ll inf=0x3f3f3f3f;
;
const ll inf_ll=(ll)1e18;
const ll maxn=100005LL;
const ll mod=1000000007LL;
+;
int arr[N];
];
int main()
{
IO
int T;
cin>>T;
while(T--){
memset(ans,,sizeof(ans));
;
cin>>n;
For(i,,n)
{
cin>>arr[i];
ans[(int )log2(arr[i])]++;
}
For(i,,n){
]={};
int x = arr[i];
while(x)
{
int x1 = x;
x&=x-;
bit[(;
}
For(j,,()
if(!bit[j])
res +=ans[j];
}
cout<<res<<endl;
}
;
}
浙江省第十二届省赛 B - Team Formation的更多相关文章
- 浙江省第十二届省赛 Beauty of Array(思维题)
Description Edward has an array A with N integers. He defines the beauty of an array as the summatio ...
- 湖南省第十二届省赛:Parenthesis
Description Bobo has a balanced parenthesis sequence P=p1 p2…pn of length n and q questions. The i-t ...
- 2015 浙江省赛B Team Formation (技巧,动归)
Team Formation For an upcoming programming contest, Edward, the headmaster of Marjar University, is ...
- HZNU第十二届校赛赛后补题
愉快的校赛翻皮水! 题解 A 温暖的签到,注意用gets #include <map> #include <set> #include <ctime> #inclu ...
- 第十二届湖南省赛G - Parenthesis (树状数组维护)
Bobo has a balanced parenthesis sequence P=p 1 p 2…p n of length n and q questions. The i-th questio ...
- 第十二届湖南省赛 (B - 有向无环图 )(拓扑排序+思维)好题
Bobo 有一个 n 个点,m 条边的有向无环图(即对于任意点 v,不存在从点 v 开始.点 v 结束的路径). 为了方便,点用 1,2,…,n 编号. 设 count(x,y) 表示点 x 到点 y ...
- 第十二届湖南省赛 A - 2016 ( 数学,同余转换)
给出正整数 n 和 m,统计满足以下条件的正整数对 (a,b) 的数量: 1. 1≤a≤n,1≤b≤m; 2. a×b 是 2016 的倍数. Input 输入包含不超过 30 ...
- Little Sub and Traveling(杭师大第十二届校赛E题) 欧拉回路
题目传送门 题目大意: 从0出发,每次只能跳到(i*2)%n或者(i*2+1)%n,求字典序最大的哈密顿回路. 思路: 首先n为奇数时无解,先来证明这一点. 先假设n为奇数,若要回到原点,则必定有一步 ...
- Little Sub and Piggybank (杭师大第十二届校赛G题) DP
题目传送门 题意:每天能往存钱罐加任意实数的钱,每天不能多于起那一天放的钱数.如果某一天的钱数恰好等于那天的特价商品,则可以买,求最后的最大快乐值. 思路:先来一段来自出题人的题解: 显然的贪心:如果 ...
随机推荐
- vsCode怎么为一个前端项目配置ts的运行环境
vsCode为一个前端项目配置ts的运行环境,ts文件保存的时候自动编译成js文件: 假设此前端项目名称为Web:文件结构如图 1. 在根目录中新建一个“.vscode”文件夹,里面建一个“tasks ...
- jQuery下拉列表二级联动插件
jQuery下拉列表二级联动插件的视图代码: <!doctype html> <html lang="en"> <head> <meta ...
- IIS 发布后无法连接数据库(应用池问题)
查找网站对应的 应用池,修改为 .net4.0 然后设置启动32位应用程序为 True
- composer应用
ubentu安装 进入自己的项目根目录cd/path/to/my/project 下载composer curl -s http://getcomposer.org/installer 把这个文件移到 ...
- Django请求原理
总结一下: 1. 进来的请求转入/hello/. 2. Django通过在ROOT_URLCONF配置来决定根URLconf. 3. Django在URLconf中的所有URL模式中,查找第一个匹配/ ...
- Python之json编码
一.json JSON: JavaScript Object Notation(JavaScript 对象表示法) JSON 是存储和交换文本信息的语法 1.json轻量级:语法规则 JSON 语法是 ...
- jquery学习之add()
解读: add() 将元素添加到匹配元素的集合中 例1: <!DOCTYPE html> <html> <head> <style> div { wid ...
- js闭包,原型,作用域等再一次理解
要理解闭包,原型等,首先要理解作用域 作用域:就是函数在定义的时候创建的,用于寻找使用到的变量的值的一个索引,而他内部的规则是,把函数自身的本地变量放在最前面,把自身的父级函数中的变量放在其次,把再高 ...
- 转:Spring-session & redis 子域名共享session
Spring-session & redis 子域名共享session 例子: a.example.com b.example.com spring 版本 4.2.6.RELEASE Spri ...
- php2go - Go 实现 PHP 常用内置函数
[转]http://www.syyong.com/Go/php2go-Use-Golang-to-implement-PHP-s-common-built-in-functions.html 使用Go ...