http://codeforces.com/contest/879/problem/C

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.

The first line contains an integer n (1 ≤ n ≤ 5·105) — 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.

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.

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

Second sample:

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.

题意：给出一段程序，只有三种操作，^ & |，然后让你输出一段不超过5行的程序使得运算结果和给出程序的运算结果相同

这题注意几点：

1.母数用000000不够，还需要与1111111结合，将其经过处理后数的每一位发生变化 0->? 1->?,

所以对于每个处理的数位我们有：

（0->0,1->0）none

（0->1,1->1）or 1

(0->1,1->0)  xor 1

(0->0,1->1)  and 1

2.位运算处理数后保持原样的操作有这三个  & 1    | 0   ^ 0

3.复习下位运算的优先级 AND>XOR>OR

#include <bits/stdc++.h>

using namespace std;

#define maxn 100000

typedef long long ll;

#define inf 2147483647

#define ri register int

int n, x;

char ch;

int f0 = , f1 = ;

int AND = , XOR = , OR = ;

int main() {

  ios::sync_with_stdio(false);

  // freopen("test.txt", "r", stdin);

  //  freopen("outout.txt","w",stdout);

  cin >> n;

  for (int i = ; i <= n; i++) {

    cin >> ch >> x;

    if (ch == '|') {

      f0 |= x;

      f1 |= x;

    }

    if (ch == '&') {

      f0 &= x;

      f1 &= x;

    }

    if (ch == '^') {

      f0 ^= x;

      f1 ^= x;

    }

  }

  int base = ;

  int w0, w1;

  for (int i = ; i <= ; i++) {

    w0 = f0 & ;

    w1 = f1 & ;

    if (w0 == ) {

      if (w1 == )

        AND += base;

    } else {

      if (w1 == ) {

        AND += base;

        XOR += base;

      } else {

        AND += base;

        OR += base;

      }

    }

    base <<= ;

    f0 >>= , f1 >>= ;

  }

  cout <<  << endl

       << "& " << AND << endl

       << "^ " << XOR << endl

       << "| " << OR << endl;

  return ;

}