Codeforces Gym 100610 Problem E. Explicit Formula 水题
Problem E. Explicit Formula
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100610
Description
Consider 10 Boolean variables x1, x2, x3, x4, x5, x6, x7, x8, x9, and x10. Consider all pairs and triplets of distinct variables among these ten. (There are 45 pairs and 120 triplets.) Count the number of pairs and triplets that contain at least one variable equal to 1. Set f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) = 1 if this number is odd and f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) = 0 if this number is even. Here’s an explicit formula that represents the function f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) correctly: f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) = (x1 ∨ x2)⊕(x1 ∨ x3)⊕(x1 ∨ x4)⊕(x1 ∨ x5)⊕(x1 ∨ x6)⊕(x1 ∨ x7)⊕ (x1 ∨ x8) ⊕ (x1 ∨ x9) ⊕ (x1 ∨ x10) ⊕ (x2 ∨ x3) ⊕ (x2 ∨ x4) ⊕ (x2 ∨ x5) ⊕ (x2 ∨ x6) ⊕ (x2 ∨ x7) ⊕ (x2 ∨ x8) ⊕ (x2 ∨ x9)⊕(x2 ∨ x10)⊕(x3 ∨ x4)⊕(x3 ∨ x5)⊕(x3 ∨ x6)⊕(x3 ∨ x7)⊕(x3 ∨ x8)⊕(x3 ∨ x9)⊕(x3 ∨ x10)⊕ (x4 ∨ x5) ⊕ (x4 ∨ x6) ⊕ (x4 ∨ x7) ⊕ (x4 ∨ x8) ⊕ (x4 ∨ x9) ⊕ (x4 ∨ x10) ⊕ (x5 ∨ x6) ⊕ (x5 ∨ x7) ⊕ (x5 ∨ x8) ⊕ (x5 ∨ x9)⊕(x5 ∨ x10)⊕(x6 ∨ x7)⊕(x6 ∨ x8)⊕(x6 ∨ x9)⊕(x6 ∨ x10)⊕(x7 ∨ x8)⊕(x7 ∨ x9)⊕(x7 ∨ x10)⊕ (x8 ∨ x9) ⊕ (x8 ∨ x10) ⊕ (x9 ∨ x10) ⊕ (x1 ∨ x2 ∨ x3) ⊕ (x1 ∨ x2 ∨ x4) ⊕ (x1 ∨ x2 ∨ x5) ⊕ (x1 ∨ x2 ∨ x6) ⊕ (x1 ∨ x2 ∨ x7) ⊕ (x1 ∨ x2 ∨ x8) ⊕ (x1 ∨ x2 ∨ x9) ⊕ (x1 ∨ x2 ∨ x10) ⊕ (x1 ∨ x3 ∨ x4) ⊕ (x1 ∨ x3 ∨ x5) ⊕ (x1 ∨ x3 ∨ x6) ⊕ (x1 ∨ x3 ∨ x7) ⊕ (x1 ∨ x3 ∨ x8) ⊕ (x1 ∨ x3 ∨ x9) ⊕ (x1 ∨ x3 ∨ x10) ⊕ (x1 ∨ x4 ∨ x5) ⊕ (x1 ∨ x4 ∨ x6) ⊕ (x1 ∨ x4 ∨ x7) ⊕ (x1 ∨ x4 ∨ x8) ⊕ (x1 ∨ x4 ∨ x9) ⊕ (x1 ∨ x4 ∨ x10) ⊕ (x1 ∨ x5 ∨ x6) ⊕ (x1 ∨ x5 ∨ x7) ⊕ (x1 ∨ x5 ∨ x8) ⊕ (x1 ∨ x5 ∨ x9) ⊕ (x1 ∨ x5 ∨ x10) ⊕ (x1 ∨ x6 ∨ x7) ⊕ (x1 ∨ x6 ∨ x8) ⊕ (x1 ∨ x6 ∨ x9) ⊕ (x1 ∨ x6 ∨ x10) ⊕ (x1 ∨ x7 ∨ x8) ⊕ (x1 ∨ x7 ∨ x9) ⊕ (x1 ∨ x7 ∨ x10) ⊕ (x1 ∨ x8 ∨ x9) ⊕ (x1 ∨ x8 ∨ x10) ⊕ (x1 ∨ x9 ∨ x10) ⊕ (x2 ∨ x3 ∨ x4) ⊕ (x2 ∨ x3 ∨ x5) ⊕ (x2 ∨ x3 ∨ x6) ⊕ (x2 ∨ x3 ∨ x7) ⊕ (x2 ∨ x3 ∨ x8) ⊕ (x2 ∨ x3 ∨ x9) ⊕ (x2 ∨ x3 ∨ x10) ⊕ (x2 ∨ x4 ∨ x5) ⊕ (x2 ∨ x4 ∨ x6) ⊕ (x2 ∨ x4 ∨ x7) ⊕ (x2 ∨ x4 ∨ x8) ⊕ (x2 ∨ x4 ∨ x9) ⊕ (x2 ∨ x4 ∨ x10) ⊕ (x2 ∨ x5 ∨ x6) ⊕ (x2 ∨ x5 ∨ x7) ⊕ (x2 ∨ x5 ∨ x8) ⊕ (x2 ∨ x5 ∨ x9) ⊕ (x2 ∨ x5 ∨ x10) ⊕ (x2 ∨ x6 ∨ x7) ⊕ (x2 ∨ x6 ∨ x8) ⊕ (x2 ∨ x6 ∨ x9) ⊕ (x2 ∨ x6 ∨ x10) ⊕ (x2 ∨ x7 ∨ x8) ⊕ (x2 ∨ x7 ∨ x9) ⊕ (x2 ∨ x7 ∨ x10) ⊕ (x2 ∨ x8 ∨ x9) ⊕ (x2 ∨ x8 ∨ x10) ⊕ (x2 ∨ x9 ∨ x10) ⊕ (x3 ∨ x4 ∨ x5) ⊕ (x3 ∨ x4 ∨ x6) ⊕ (x3 ∨ x4 ∨ x7) ⊕ (x3 ∨ x4 ∨ x8) ⊕ (x3 ∨ x4 ∨ x9) ⊕ (x3 ∨ x4 ∨ x10) ⊕ (x3 ∨ x5 ∨ x6) ⊕ (x3 ∨ x5 ∨ x7) ⊕ (x3 ∨ x5 ∨ x8) ⊕ (x3 ∨ x5 ∨ x9) ⊕ (x3 ∨ x5 ∨ x10) ⊕ (x3 ∨ x6 ∨ x7) ⊕ (x3 ∨ x6 ∨ x8) ⊕ (x3 ∨ x6 ∨ x9) ⊕ (x3 ∨ x6 ∨ x10) ⊕ (x3 ∨ x7 ∨ x8) ⊕ (x3 ∨ x7 ∨ x9) ⊕ (x3 ∨ x7 ∨ x10) ⊕ (x3 ∨ x8 ∨ x9) ⊕ (x3 ∨ x8 ∨ x10) ⊕ (x3 ∨ x9 ∨ x10) ⊕ (x4 ∨ x5 ∨ x6) ⊕ (x4 ∨ x5 ∨ x7) ⊕ (x4 ∨ x5 ∨ x8) ⊕ (x4 ∨ x5 ∨ x9) ⊕ (x4 ∨ x5 ∨ x10) ⊕ (x4 ∨ x6 ∨ x7) ⊕ (x4 ∨ x6 ∨ x8) ⊕ (x4 ∨ x6 ∨ x9) ⊕ (x4 ∨ x6 ∨ x10) ⊕ (x4 ∨ x7 ∨ x8) ⊕ (x4 ∨ x7 ∨ x9) ⊕ (x4 ∨ x7 ∨ x10) ⊕ (x4 ∨ x8 ∨ x9) ⊕ (x4 ∨ x8 ∨ x10) ⊕ (x4 ∨ x9 ∨ x10) ⊕ (x5 ∨ x6 ∨ x7) ⊕ (x5 ∨ x6 ∨ x8) ⊕ (x5 ∨ x6 ∨ x9) ⊕ (x5 ∨ x6 ∨ x10) ⊕ (x5 ∨ x7 ∨ x8) ⊕ (x5 ∨ x7 ∨ x9) ⊕ (x5 ∨ x7 ∨ x10) ⊕ (x5 ∨ x8 ∨ x9) ⊕ (x5 ∨ x8 ∨ x10) ⊕ (x5 ∨ x9 ∨ x10) ⊕ (x6 ∨ x7 ∨ x8) ⊕ (x6 ∨ x7 ∨ x9) ⊕ (x6 ∨ x7 ∨ x10) ⊕ (x6 ∨ x8 ∨ x9) ⊕ (x6 ∨ x8 ∨ x10) ⊕ (x6 ∨ x9 ∨ x10) ⊕ (x7 ∨ x8 ∨ x9) ⊕ (x7 ∨ x8 ∨ x10) ⊕ (x7 ∨ x9 ∨ x10) ⊕ (x8 ∨ x9 ∨ x10) In this formula ∨ stands for logical or, and ⊕ stands for exclusive or (xor). Remember that in C++ and Java these two binary operators are denoted as “||” and “^”. Given the values of x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, calculate the value of f(x1, x2, . . . , x10).
Input
The input file contains 10 numbers x1, x2, x3, x4, x5, x6, x7, x8, x9, and x10. Each of them is either 0 or 1.
Output
Output a single value — f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10).
Sample Input
1 0 0 1 0 0 1 0 0 1
Sample Output
0
HINT
题意
就求题目给的那个式子的答案是多少
题解:
ctrl+f 把符号替换一下就好了……
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 110000
#define mod 10007
#define eps 1e-9
#define pi 3.1415926
int Num;
//const int inf=0x7fffffff; //§ß§é§à§é¨f§³
const ll Inf=0x3f3f3f3f3f3f3f3fll;
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;
}
//************************************************************************************* ll x1,x2,x3,x4,x5,x6,x7,x8,x9,x10;
int main()
{
freopen("explicit.in","r",stdin);
freopen("explicit.out","w",stdout);
cin>>x1>>x2>>x3>>x4>>x5>>x6>>x7>>x8>>x9>>x10; cout<<((x1 || x2)^(x1 || x3)^(x1 || x4)^(x1 || x5)^(x1 || x6)^(x1 || x7)^
(x1 || x8) ^ (x1 || x9) ^ (x1 || x10) ^ (x2 || x3) ^ (x2 || x4) ^ (x2 || x5) ^ (x2 || x6) ^ (x2 || x7) ^ (x2 || x8) ^
(x2 || x9)^(x2 || x10)^(x3 || x4)^(x3 || x5)^(x3 || x6)^(x3 || x7)^(x3 || x8)^(x3 || x9)^(x3 || x10)^
(x4 || x5) ^(x4 || x6) ^ (x4 || x7) ^ (x4 || x8) ^ (x4 || x9) ^ (x4 || x10) ^ (x5 || x6) ^ (x5 || x7) ^ (x5 || x8) ^
(x5 || x9)^(x5 || x10)^(x6 || x7)^(x6 || x8)^(x6 || x9)^(x6 || x10)^(x7 || x8)^(x7 || x9)^(x7 || x10)^
(x8 || x9) ^ (x8 || x10) ^(x9 || x10) ^ (x1 || x2 || x3) ^ (x1 || x2 || x4) ^ (x1 || x2 || x5) ^ (x1 || x2 || x6) ^
(x1 || x2 || x7) ^ (x1 || x2 || x8) ^ (x1 || x2 || x9) ^ (x1 || x2 || x10) ^ (x1 || x3 || x4) ^ (x1 || x3 || x5) ^
(x1 || x3 || x6) ^ (x1 || x3 || x7) ^ (x1 || x3 || x8) ^ (x1 || x3 || x9) ^ (x1 || x3 || x10) ^ (x1 || x4 || x5) ^
(x1 || x4 || x6) ^ (x1 || x4 || x7) ^ (x1 || x4 || x8) ^ (x1 || x4 || x9) ^ (x1 || x4 || x10) ^ (x1 || x5 || x6) ^
(x1 || x5 || x7) ^ (x1 || x5 || x8) ^ (x1 || x5 || x9) ^ (x1 || x5 || x10) ^ (x1 || x6 || x7) ^ (x1 || x6 || x8) ^
(x1 || x6 || x9) ^ (x1 || x6 || x10) ^ (x1 || x7 || x8) ^ (x1 || x7 || x9) ^ (x1 || x7 || x10) ^ (x1 || x8 || x9) ^
(x1 || x8 || x10) ^ (x1 || x9 || x10) ^ (x2 || x3 || x4) ^ (x2 || x3 || x5) ^ (x2 || x3 || x6) ^ (x2 || x3 || x7) ^
(x2 || x3 || x8) ^ (x2 || x3 || x9) ^ (x2 || x3 || x10) ^ (x2 || x4 || x5) ^ (x2 || x4 || x6) ^ (x2 || x4 || x7) ^
(x2 || x4 || x8) ^ (x2 || x4 || x9) ^ (x2 || x4 || x10) ^ (x2 || x5 || x6) ^ (x2 || x5 || x7) ^ (x2 || x5 || x8) ^
(x2 || x5 || x9) ^ (x2 || x5 || x10) ^ (x2 || x6 || x7) ^ (x2 || x6 || x8) ^ (x2 || x6 || x9) ^ (x2 || x6 || x10) ^
(x2 || x7 || x8) ^ (x2 || x7 || x9) ^ (x2 || x7 || x10) ^ (x2 || x8 || x9) ^ (x2 || x8 || x10) ^ (x2 || x9 || x10) ^
(x3 || x4 || x5) ^ (x3 || x4 || x6) ^ (x3 || x4 || x7) ^ (x3 || x4 || x8) ^ (x3 || x4 || x9) ^ (x3 || x4 || x10) ^
(x3 || x5 || x6) ^ (x3 || x5 || x7) ^ (x3 || x5 || x8) ^ (x3 || x5 || x9) ^ (x3 || x5 || x10) ^ (x3 || x6 || x7) ^
(x3 || x6 || x8) ^ (x3 || x6 || x9) ^ (x3 || x6 || x10) ^ (x3 || x7 || x8) ^ (x3 || x7 || x9) ^ (x3 || x7 || x10) ^
(x3 || x8 || x9) ^ (x3 || x8 || x10) ^ (x3 || x9 || x10) ^ (x4 || x5 || x6) ^ (x4 || x5 || x7) ^ (x4 || x5 || x8) ^
(x4 || x5 || x9) ^ (x4 || x5 || x10) ^ (x4 || x6 || x7) ^ (x4 || x6 || x8) ^ (x4 || x6 || x9) ^ (x4 || x6 || x10) ^
(x4 || x7 || x8) ^ (x4 || x7 || x9) ^ (x4 || x7 || x10) ^ (x4 || x8 || x9) ^ (x4 || x8 || x10) ^ (x4 || x9 || x10) ^
(x5 || x6 || x7) ^ (x5 || x6 || x8) ^ (x5 || x6 || x9) ^ (x5 || x6 || x10) ^ (x5 || x7 || x8) ^ (x5 || x7 || x9) ^
(x5 || x7 || x10) ^ (x5 || x8 || x9) ^ (x5 || x8 || x10) ^ (x5 || x9 || x10) ^ (x6 || x7 || x8) ^ (x6 || x7 || x9) ^
(x6 || x7 || x10) ^ (x6 || x8 || x9) ^ (x6 || x8 || x10) ^ (x6 || x9 || x10) ^ (x7 || x8 || x9) ^ (x7 || x8 || x10) ^
(x7 || x9 || x10) ^ (x8 || x9 || x10)) <<endl; }
Codeforces Gym 100610 Problem E. Explicit Formula 水题的更多相关文章
- Codeforces Gym 100610 Problem A. Alien Communication Masterclass 构造
Problem A. Alien Communication Masterclass Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codefo ...
- Codeforces Gym 100610 Problem K. Kitchen Robot 状压DP
Problem K. Kitchen Robot Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10061 ...
- Codeforces Gym 100610 Problem H. Horrible Truth 瞎搞
Problem H. Horrible Truth Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/1006 ...
- codeforces Gym 100187L L. Ministry of Truth 水题
L. Ministry of Truth Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/p ...
- Codeforces Gym 100342H Problem H. Hard Test 构造题,卡迪杰斯特拉
Problem H. Hard TestTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/at ...
- Codeforces Gym 100523C C - Will It Stop? 水题
C - Will It Stop?Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...
- Codeforces Round #185 (Div. 2) B. Archer 水题
B. Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/B D ...
- Gym 101873K - You Are Fired - [贪心水题]
题目链接:http://codeforces.com/gym/101873/problem/K 题意: 现在给出 $n(1 \le n \le 1e4)$ 个员工,最多可以裁员 $k$ 人,名字为 $ ...
- Educational Codeforces Round 14 A. Fashion in Berland 水题
A. Fashion in Berland 题目连接: http://www.codeforces.com/contest/691/problem/A Description According to ...
随机推荐
- TCP/IP详解学习笔记(12)-TCP的超时与重传
超时重传是TCP协议保证数据可靠性的另一个重要机制,其原理是在发送某一个数据以后就开启一个计时器,在一定时间内如果没有得到发送的数据报的ACK报文,那么就重新发送数据,直到发送成功为止. 1.超时 超 ...
- shell 删除日志
一般线上服务的日志都是采用回滚的防止,写一定数量的日志 或是有管理工具定期去转移老旧日志 前几天删除一个测试环境的日志,只保留两天的日志,结果把正在写的日志都给删掉了,不得不重启了服务,经过这一次的错 ...
- redhat--nagios插件--check_traffic.sh
****在被监控主机安装nrpe**** (1)在被监控主机上,增加用户和密码 useradd nagios passwd nagios (2)安装nagios插件 tar zxf nagios-pl ...
- IIS应用程序池回收图文详解
转:http://blog.sina.com.cn/s/blog_8677fcaa010138uf.html 什么是应用程序池呢?这是微软的一个全新概念:应用程序池是将一个或多个应用程序链接到一个或多 ...
- memmove函数
写一个函数,完成内存之间的拷贝 void* mymemcpy( void *dest, const void *src, size_t count ) { char* pdest = static_c ...
- 数组乘积--满足result[i] = input数组中除了input[i]之外所有数的乘积(假设不会溢出
数组乘积(15分) 输入:一个长度为n的整数数组input 输出:一个长度为n的整数数组result,满足result[i] = input数组中除了input[i]之外所有数的乘积(假设不会溢出). ...
- [转]解决crystal report水晶报表在浏览器提示bobj未定义的错误
网上的中文文章(比如这篇文章)都是写的部署到服务器后出现的问题,同时也指出要把crystal report的aspnet_client文件夹拷贝到对应项目的根目录里,这样就可以正常显示了,但是具体到我 ...
- IOS AsyncSocket
导入AsyncSocket.h AsyncSocket.m AsyncUdpSocket.h AsyncUdpSocket.m 以及 CFNetWork.framework async ...
- 给Webkit内核的浏览器控件增加互交功能
转载请说明出处,谢谢~~ 昨天封装了基于webkit的wke浏览器内核,做成了duilib的浏览器控件,实现了浏览功能,但是单单的浏览功能还不满足需求,在我的仿酷狗项目中乐库的功能需要与浏览器互交. ...
- opencv行人检测里遇到的setSVMDetector()问题
参考了博客http://blog.csdn.net/carson2005/article/details/7841443 后,自己动手后发现了一些问题,博客里提到的一些问题没有解决 ,是关于为什么图像 ...