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:

 1 int GCD_2(int nNum1, int nNum2)
 2 {
 3     if (nNum1 > nNum2)
 4     {
 5         nNum1 = nNum1 ^ nNum2;
 6         nNum2 = nNum1 ^ nNum2;
 7         nNum1 = nNum1 ^ nNum2;
 8     }
 9     if (nNum1 == 0)
10     {
11         return nNum2;
12     }
13     return GCD_2(nNum2 % nNum1, nNum1);
14 }

遍历 vector ,迭代 GCD_2 ,计算N个数的最大公因数:

1 int GCD_N(std::vector<int> vecNums)
2 {
3     int nRes = vecNums[0];
4     std::for_each(vecNums.begin() + 1, vecNums.end(),
5                   [&nRes](int x){ nRes = GCD_2(nRes, x); });
6     return nRes;
7 }

测试:

1 int main()
2 {
3     int nArr[] = { 75, 45, 315 };
4     std::vector<int> vecNums(nArr, nArr + sizeof(nArr) / 4);
5     int nRes = GCD_N(vecNums);
6     std::cout << nRes << std::endl;
7     return 0;
8 }

结果:

1 15

 

posted @ 2015-02-24 13:14  ZChameleon  阅读(278)  评论(0)  编辑  收藏  举报