Codeforces Round #443 (Div. 1) A. Short Program
A. Short Program
link
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
input
3
| 3
^ 2
| 1
output
2
| 3
^ 2
input
3
& 1
& 3
& 5
output
1
& 1
input
3
^ 1
^ 2
^ 3
output
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.
翻译
现在有一个程序，输入一个[0,1024)的数，然后经过一些位运算之后，输出一个数。
现在让你简化中间的位运算的过程，使得不超过5步，使得答案和原程序一样。
题解
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。
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;
}最新文章
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