二进制反射格雷码

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#include <iostream>#include <cmath>#include <vector>#include <string>using namespace std;#define MAX 1000int BCD[MAX];int GRAYCODE[MAX];int temp[MAX];//BCD码转换成格雷码void bcdToGraycode(int bcd[], int gray[], int rightIndex){if (rightIndex == 0){gray[rightIndex] = bcd[rightIndex];return;}else{gray[rightIndex] = bcd[rightIndex] ^ bcd[rightIndex - 1];bcdToGraycode(bcd, gray, rightIndex - 1);}}void generaterGrayCode(int n){vector<string> grayCodeVec;//当n为1的时候的格雷码  string aa = "0";string bb = "1";grayCodeVec.push_back(aa);grayCodeVec.push_back(bb);//产生大于两位的格雷码，n位格雷码的数量为2^n个  if (n > 1){for (int i = 2; i <= n; i++){//设置一个临时存储空间来存储n-1位格雷码  vector<string> tempGrayCodeVec;for (size_t k = 0; k < grayCodeVec.size(); k++){tempGrayCodeVec.push_back(grayCodeVec[k]);}//在前面产生的n-1位格雷码前面添加一位数0产生2^(n-1)个n位格雷码，并替换掉原来的n-1位格雷码  int tempGrayCodeVecSize = tempGrayCodeVec.size();for (int j = 0; j < tempGrayCodeVecSize; j++){string tempbitzero = "0";tempbitzero += tempGrayCodeVec[j];grayCodeVec[j] = tempbitzero;}//将前面产生的n-1位格雷码的顺序反转  //在反转后的n-1位格雷码前面添加一位数1产生剩下2^(n-1)个n位格雷码，并存储起来  for (int jj = tempGrayCodeVecSize - 1; jj >= 0; jj--){string tempbitone = "1";tempbitone += tempGrayCodeVec[jj];grayCodeVec.push_back(tempbitone);}//释放掉临时存储空间  tempGrayCodeVec.clear();}}//输出n位格雷码  for (size_t i = 0; i < grayCodeVec.size(); i++){cout << grayCodeVec[i] << endl;}cout << endl;}int main(){int number;//二进制数组长度int index = 0;int len;int m = 2;int t;//输入一个十进制数cin >> number;//转换成二进制数while (number){temp[index++] = number % 2;number = number >> 1;}//按照标准格式输出，位数为2的整数次幂while (1){t = pow(2, m);if (index <= t){len = t - 1;break;}else{m++;}}for (int i = 0; i <= len; i++){BCD[i] = 0;}t = index - 1;for (int i = 0; i < index; i++)BCD[len - index + 1 + i] = temp[t--];bcdToGraycode(BCD, GRAYCODE, len);for (int i = 0; i <= len; i++)cout << GRAYCODE[i] << ' ';generaterGrayCode(4);system("pause");return 0;}

40 1 1 0 0000000100110010011001110101010011001101111111101010101110011000请按任意键继续. . .

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