求N个数的最大公约数

使用 “辗转相除法” 计算2个数的最大公因数:

 int GCD_2(int nNum1, int nNum2)
 {
     if (nNum1 > nNum2)
     {
         nNum1 = nNum1 ^ nNum2;
         nNum2 = nNum1 ^ nNum2;
         nNum1 = nNum1 ^ nNum2;
     }
     if (nNum1 == )
     {
         return nNum2;
     }
     return GCD_2(nNum2 % nNum1, nNum1);
 }

遍历 vector ,迭代 GCD_2 ,计算N个数的最大公因数:

 int GCD_N(std::vector<int> vecNums)
 {
     int nRes = vecNums[];
     std::for_each(vecNums.begin() + , vecNums.end(),
                   [&nRes](int x){ nRes = GCD_2(nRes, x); });
     return nRes;
 }

测试:

 int main()
 {
     int nArr[] = { , ,  };
     std::vector<int> vecNums(nArr, nArr + sizeof(nArr) / );
     int nRes = GCD_N(vecNums);
     std::cout << nRes << std::endl;
     return ;
 }

结果: