Codeforces Round #443 (Div. 2)
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:给一个任意数x,进行位运算,求怎么简化到不超过5次。
分析:看出每次的操作数不超过2^10-1,先用0000000000,1111111111,进行题意的操作,发现规律,01-----> 0/1,通过 | ^ & 运算使得它成立。
#include <bits/stdc++.h> using namespace std; bool calc(int a,int i) {
if(a&(<<i)) return ;
return ;
} int main()
{
int n;
int x = ,y = ; cin>>n;
while(n--) {
char str[];
int t;
scanf("%s%d",str,&t); if(str[]=='|') x|=t,y|=t; if(str[]=='&') x&=t,y&=t; if(str[]=='^') x^=t,y^=t; } int v1 = ; // |
int v2 = ; // ^
int v3 = ;
for(int i = ; i < ; i++) {
if(calc(x,i)&&calc(y,i)) v1 = v1 + (<<i);
if(calc(x,i)&&!calc(y,i)) v2 = v2 + (<<i);
//if(!calc(x,i)&&calc(y,i))
if(!calc(x,i)&&!calc(y,i)) v3 = v3 - (<<i);
} printf("3\n"); printf("| %d\n",v1);
printf("^ %d\n",v2);
printf("& %d\n",v3); return ;
}
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