思路一：给定a，b两个数，将较小的数赋值给c，令c=b（不妨设a>b），将a和b对c做求余运算，从c开始依次减1向下遍历，直到两者的余数都为0则返回c，否则继续循环遍历。

代码如下：

#include<iostream.h>

int gcd(int a,int b)

{int c;

c=(a>b)?b:a;

while(a%c!=0||b%c!=0)

{c--;

}

return c;

}

int main()

{int a,b,max_div;

cout<<"please input two integer:";

cin>>a>>b;

max_div=gcd(a,b);

cout<<"the greatest common divisor of ";

cout<<a<<" and "<<b<<" is "<<max_div<<endl;

return 0;

}

思路二：辗转相除法

假设a>b,如果a不能被b整除，则将b赋值给a，余数赋值给b，重复执行a%b,直到a能够被b整除。此时返回b的值，则为最大公约数。

代码如下：

#include<iostream.h>

int gcd2(int a,int b)

{int c;

if(a<b)

{a=a+b;

b=a-b;

a=a-b;

}

c=a%b;

while(a%b!=0)

{a=b;

b=c;

c=a%b;

}

return b;

}

int main()

{int a,b,max_div;

cout<<"please input two integer:";

cin>>a>>b;

max_div=gcd2(a,b);

cout<<"the greatest common divisor of ";

cout<<a<<" and "<<b<<" is "<<max_div<<endl;

return 0;

}