给予多个正整数，然后求它们的最大公约数gcd？

#include <iostream> #include <vector> int gcd(std::vector<int>* pivec); int main(void) {         std::vector<int> ivec;         int a;         while (std::cin >> a)                 ivec.push_back(a);         for (std::vector<int>::size_type i = 0; i != ivec.size(); ++i)                 std::cout << ivec[i] << std::endl;         std::cout << gcd(&ivec) << std::endl;         for (std::vector<int>::size_type i = 0; i != ivec.size(); ++i)                 std::cout << ivec[i] << std::endl;         return 0; } int gcd(std::vector<int>* pivec) {         int smallest = (*pivec)[0];         //make sure smallest is the smallest in the vector<int>         for (std::vector<int>::size_type i = 1; i != pivec->size(); ++i)                 if (smallest > (*pivec)[i])                         smallest = (*pivec)[i];         //历遍2~smallest         int gcd = 1;         for (int i = 2; i != smallest + 1; ++i) {                 bool flag = 1;         //只有不跳出来，就说明对于这个i，vector当中每一个数都可以被这个i整除，i是我要的         for (std::vector<int>::iterator iter = pivec->begin(); iter != pivec->end(); ++iter)                 if (*iter % i != 0) {                         flag = 0;                         break;                 }         if (flag == 1) {                 gcd *= i;                 for (std::vector<int>::iterator iter = pivec->begin(); iter != pivec->end(); ++iter)                         *iter /= i;                 smallest /= i;                 i = 1;         }         }         return gcd; }

以上的代码其实很烦人。而且将原始输入的vector都给改变了。不划算
但是这种方法有一个好处，能够得到质因数。

还有一个方法，不错，很简单。灵感来源于
https://www.tutorialspoint.com/cplusplus-program-for-gcd-of-more-than-two-or-array-numbers

#include <iostream> #include <vector> int gcd(int a, int b); int ngcd(std::vector<int>* pivec); int main(void) {         std::vector<int> ivec;         int a;         while (std::cin >> a)                 ivec.push_back(a);         for (std::vector<int>::size_type i = 0; i != ivec.size(); ++i)                 std::cout << ivec[i] << std::endl;         std::cout << ngcd(&ivec) << std::endl;         for (std::vector<int>::size_type i = 0; i != ivec.size(); ++i)                 std::cout << ivec[i] << std::endl;         return 0; } int ngcd(std::vector<int>* pivec) {         int r = (*pivec)[0];         for (std::vector<int>::size_type i = 1; i != pivec->size(); ++i)                 r = gcd(r, (*pivec)[i]);         return         r; } int gcd(int a, int b) {         while (a != b)                 (a < b ? b : a) = std::abs(a - b);         return a; }

坏处显而易见，没有质因数。

以上两种方法，一运行，什么都不输入，直接ctrl+d，会segmentation fault。