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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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/ ...

  7. 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 ...

  8. 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 $ ...

  9. 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

  1. 【Unity3D】Unity自带组件—完成第一人称人物控制

    1.导入unity自带的Character Controllers包 2.可以看到First Person Controller组件的构成 Mouse Look() : 随鼠标的移动而使所属物体发生旋 ...

  2. 也谈http中get和post

    1.get和post区别: 从设计初衷考虑get是为了查询服务器资源(不改变服务器数据及状态,因此说它是安全和幂等的,但get请求参数一般是直接在url后面,浏览器地址栏中会被看到能保存书签及历史记录 ...

  3. [HTML Q&A][转]使pre的内容自动换行

    <pre> 元素可定义预格式化的文本.被包围在 pre 元素中的文本通常会保留空格和换行符.而文本也会呈现为等宽字体. <pre> 标签的一个常见应用就是用来表示计算机的源代码 ...

  4. Ubuntu 出现 apt-get问题的解决方法

    ubuntu 10.10 sudo apt-get update 404  Not Found or W: Failed to fetch http://cn.old-releases.ubuntu. ...

  5. 两个数组a[N],b[N],其中A[N]的各个元素值已知,现给b[i]赋值,b[i] = a[0]*a[1]*a[2]…*a[N-1]/a[i];

    转自:http://blog.csdn.net/shandianling/article/details/8785269 问题描述:两个数组a[N],b[N],其中A[N]的各个元素值已知,现给b[i ...

  6. 【LeetCode 173】Binary Search Tree Iterator

    Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the ro ...

  7. Linux(CentOs)下安装Phantomjs + Casperjs

    Linux(CentOs)下安装Phantomjs + Casperjs 是参照cnMiss's Blog http://ju.outofmemory.cn/entry/70691的博客进行安装的 1 ...

  8. (转载) VS编译duilib项目时候的错误解决方法整理

    原文地址:http://blog.csdn.net/x356982611/article/details/30217473 @1:找不到Riched20.lib 用everything等软件搜索下磁盘 ...

  9. 理解CSS盒子模型

    概述 网页设计中常听的属性名:内容(content).填充(padding).边框(border).边界(margin),CSS盒子模型都具备这些属性,也主要是这些属性. 这些属性我们可以把它转移到我 ...

  10. 路径 (Path)–nodejs

    本模块包含一套用于处理和转换文件路径的工具集.几乎所有的方法只做字符串变换, 不会调用文件系统检查路径是否有效. 通过 require('path') 来加载此模块.以下是本模块所提供的方法: pat ...