标签：logs upper not c++ second ant this cal lan
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&(1<<i)) return 1; return 0; } int main() { int n; int x = 0,y = 1023; cin>>n; while(n--) { char str[5]; int t; scanf("%s%d",str,&t); if(str[0]==‘|‘) x|=t,y|=t; if(str[0]==‘&‘) x&=t,y&=t; if(str[0]==‘^‘) x^=t,y^=t; } int v1 = 0; // | int v2 = 0; // ^ int v3 = 1023; for(int i = 0; i < 10; i++) { if(calc(x,i)&&calc(y,i)) v1 = v1 + (1<<i); if(calc(x,i)&&!calc(y,i)) v2 = v2 + (1<<i); //if(!calc(x,i)&&calc(y,i)) if(!calc(x,i)&&!calc(y,i)) v3 = v3 - (1<<i); } printf("3\n"); printf("| %d\n",v1); printf("^ %d\n",v2); printf("& %d\n",v3); return 0; }
Codeforces Round #443 (Div. 2)
标签：logs upper not c++ second ant this cal lan
原文地址：http://www.cnblogs.com/TreeDream/p/7751329.html
