Short Program CodeForces - 878A

最新推荐文章于 2023-12-02 22:33:12 发布

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本文介绍了一个关于CALPAS编程语言的挑战问题：如何用不超过5行代码实现与给定程序相同的功能。输入包含一系列位操作指令，目标是简化这些指令，确保对于0到1023的所有输入都能得到相同的结果。

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.

Example
Input
3
| 3
^ 2
| 1
Output
2
| 3
^ 2
Input
3
& 1
& 3
& 5
Output
1
& 1
Input
3
^ 1
^ 2
^ 3
Output
0
这题输出的时候忘了加空格啊，，，

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
bool check(int x,int i)
{
    return x&(1<<i);
}
int main()
{
    int n;
    int x1=0;
    int x2=1023;
    char op[5];
    int num;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%s%d",op,&num);
        if(op[0]=='|')
            x1|=num,x2|=num;
        else
            if(op[0]=='^')
                x1^=num,x2^=num;
            else
                x1&=num,x2&=num;
    }

    int v1=0,v2=0;
    for(int i=0;i<10;i++)
    {
        if(check(x1,i)&&check(x2,i))
            v2+=(1<<i);
        else
            if(check(x1,i)&&!check(x2,i))
                v1+=(1<<i),v2+=(1<<i);
            else
                if(!check(x1,i)&&check(x2,i))
                    v1+=(1<<i);
    }
    printf("2\n");
    cout<<"& "<<v1<<endl;
    cout<<"^ "<<v2<<endl;
    return 0;
}