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

## A. Short Program

http://codeforces.com/contest/878/problem/A

## describe

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.

## Examples

3
| 3
^ 2
| 1

2
| 3
^ 2

3
& 1
& 3
& 5

1
& 1

3
^ 1
^ 2
^ 3

0

### Note

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.

(1)0->0,1->0
(2)0->1,1->0
(3)0->0,1->1
(4)0->1,1->1

## 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=='|'){
a|=p;
b|=p;
}else if(s=='^'){
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;
}
posted @ 2017-10-30 19:12 qscqesze 阅读(...) 评论(...) 编辑 收藏