Codeforces Round #879 (Div. 2) C. Short Program

题目链接：http://codeforces.com/contest/879/problem/C

C. Short Program

time limit per test2 seconds

memory limit per test256 megabytes

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.

Input

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

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.

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别人写的太好了，不敢下手。

别人家的博客

之前题意了解错了，就是位运算的等价变换（看不懂的可以试试将位运算替换成加减）。

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#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b;
    a = 0,b = 1023;
    int n;
    scanf("%d",&n);
    char s[100];
    int num;
    while(n--)
    {
        scanf("%s%d",s,&num);
        if(s[0] == '|')
        {
            a |= num;
            b |= num;
        }
        else if(s[0] == '&')
        {
            a &= num;
            b &= num;
        }
        else if(s[0] == '^')
        {
            a ^= num;
            b ^= num;
        }
    }
    int num_and = 0,num_or = 0,num_xor = 0;
    num_and = a | b;//0->0,1->1，可以与上b二进制表示中1的部分
    num_or = a & b;//0->1,1->1，两个二进制中都是1的部分
    num_xor = a & (b ^ 1023);//0->1,1->0，两个二进制中都变成1的部分
    printf("3\n");
    printf("| %d\n",num_or);
    printf("& %d\n",num_and);
    printf("^ %d\n",num_xor);
    return 0;
}