C++入门程序作业3

/*

输出n位数据的格雷码
The gray code is a binary numeral system where two successive values differ in only one bit.
Given a non-negative integer n representing the total number of bits in the code, print the sequence of
gray code. A gray code sequence must begin with 0.
For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:
00 - 0
01 - 1
11 - 3
10 - 2
Note:
For a given n, a gray code sequence is not uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.
For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.

要求用向量做这道题，不能用公式
*/
#include<math.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;

 

int main()
{
vector<int>num0;//if n is determined, the number need to be transformed
vector<int>num1;//from nomal number to 2 bit vector
vector<int>num2;//to grey
vector<int>num3;//to normal-binary
vector<int>num4;//to align the grey
int j=0;
//according to n generate part of number;
int n=4;

for (int i=0;i<pow(2,n);i++)
num0.push_back(i);

cout<<"Given n = "<<n<<" , return [ ";
for (int i=0;i<num0.size()-1;i++)
cout << num0[i] << ", ";
cout << num0[num0.size()-1];
cout<<" ].";
cout << endl;

for(j=0;j< pow(2,n);j++) {

vector<int>vnum;
int fnum=0;

fnum=num0[j];

int num=fnum;
int i=0;

for(i=0;num>2;i++)//frist push LSB
{

vnum.push_back(num%2);
num=num/2;
}
vnum.push_back(num%2);

int a=1;

num1.push_back(0);//add 0 in the MSB
for (i=1;i<=vnum.size();i++)
num1.push_back(vnum[vnum.size()-i]);

//generate gray code in binary
for (int i=0;i<num1.size()-1;i++)
num2.push_back(num1[i]^num1[i+1] );

//generate gray code in align
num4=num2;

int s=0;
int k=0;
s=n-num1.size();

for(k=0;k<=s;k++){
num4.push_back(0);
}

//grey to normal-binary
int c=0^num2[0];
num3.push_back(c);
for (int i=1;i<num2.size();i++)
{
c=c^num2[i];
num3.push_back(c);

}

//display

for (int i=0;i<num4.size();i++)
cout << num4[i] << " ";
cout <<" - "<<fnum<< endl;
num1.clear( );
num2.clear( );
num3.clear( );
num4.clear( );
}

return 0;
}