Codeforces Round #443 (Div. 2) C. Short Program
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.
3
| 3
^ 2
| 1
2
| 3
^ 2
3
& 1
& 3
& 5
1
& 1
3
^ 1
^ 2
^ 3
0
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 思路:经过如果操作之后,总共十位二进制数,有:确定为1的,确定为0的,不变的,是原来的相反的,然后分别讨论四种 情况 */ #include <bits/stdc++.h> #define MAXN 500005 #define MAXK 2 using namespace std; int n; char str[MAXN][MAXK]; int a[MAXN]; int ora,xora,anda; inline int cal(int x){ for(int i=0;i<n;i++){ switch(str[i][0]){ case '^': x^=a[i]; break; case '&': x&=a[i]; break; case '|': x|=a[i]; break; } } return x; } inline void init(){ ora=0; xora=0; anda=0; } int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%s%d",str[i],&a[i]); } int pos=cal(0); for(int i=0;i<10;i++){ int cur=cal((1<<i)); if(( cur&(1<<i) )==0&&( pos&(1<<i) )==0){//如果这一位肯定为0 }else if(( cur&(1<<i) )!=0&&( pos&(1<<i) )!=0){//这位肯定为1 anda+=(1<<i); ora+=(1<<i); }else if(( cur&(1<<i) )==0&&( pos&(1<<i) )!=0){//这位和原来相反 anda+=(1<<i); xora+=(1<<i); }else{//这位不变 anda+=(1<<i); } } printf("3\n"); printf("& %d\n",anda); printf("| %d\n",ora); printf("^ %d\n",xora); return 0; }