Codeforces Round #443 (Div. 1) A. Short Program

时间限制：1S / 空间限制：256MB

【在线测试提交传送门】

【问题描述】

    Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well.In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned.Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya's program, and consists of no more than 5 lines. Your program should return the same integer as Petya's program for all arguments from 0 to 1023.  该程序设计语言一共能执行三种运算，AND， OR ， XOR ，通过一系列运算可以将[0,1023]范围内的整数转换为另外一个整数。现在给出一些转换程序语句，请你将语句缩减为不超过5行，得到和原程序同样的输出。

【输入格式】

The first line contains an integer n (1 ≤ n ≤ 5·10^5) — the number of lines.Next n lines contain commands. A command consists of a character that represents the operation ("&", "|" or "^" for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023.第一行，一个整数n (1 ≤ n ≤ 5·10^5) ，表示行数。接下来n行，每行包含一条命令：一条命令用一个字符代表位运算，分别是"&", "|" or "^"，即与、或、抑或，一个常数xi（0 ≤ xi ≤ 1023）。

【输出格式】

Output an integer k (0 ≤ k ≤ 5) — the length of your program.Next k lines must contain commands in the same format as in the input.输出包含一个整数K(0 ≤ k ≤ 5) 行，表示程序的的行数；接下来k行，每行包含一条命令，格式与输入一致。

【输入样例1】

3| 3^ 2| 1

【输出样例1】

2| 3^ 2

【输入样例2】

3& 1& 3& 5

【输出样例2】

1& 1

【样例2解释】

Let x be an input of the Petya's program. It's output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs.

【输入样例3】

3^ 1^ 2^ 3

【输出样例3】

0

【提示】

You can read about bitwise operations in https://en.wikipedia.org/wiki/Bitwise_operation.

【解题思路】

考虑每一位，只会存在4种情况：
对于输入的数字的每一位而言，要么是1，要么是0。而这些每一位的数输出之后要么变成了0，要么就变成了1.
(1)0->0,1->0
(2)0->1,1->0
(3)0->0,1->1
(4)0->1,1->1
对于四种情况，我们都可以通过^和|就可以解决，分情况讨论输出即可。

【参考代码】

#include<bits/stdc++.h>using namespace std;int n,a,b,p;string s;int main(){    cin>>n;    a = 0,b = 1023;    for(int i=0;i<n;i++){        cin>>s>>p;        if(s[0]=='|'){            a|=p;            b|=p;        }else if(s[0]=='^'){            a^=p;            b^=p;        }else{            a&=p;            b&=p;        }    }    int ans1=0,ans2=0;    for(int i=0;i<10;i++){        int a1=a&(1<<i);        int b1=b&(1<<i);        if(a1&&b1){            ans1|=(1<<i);        }        if(a1&&!b1){            ans2|=(1<<i);        }        if(!a1&&!b1){            ans1|=(1<<i);            ans2|=(1<<i);        }    }    cout<<"2"<<endl;    cout<<"| "<<ans1<<endl;    cout<<"^ "<<ans2<<endl;}