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求两个数的最大公因子，使用“辗转相除法”。
原理如下：若r=a%b，则gcd(a,b)=gcd(b,r)。
简单推导：
因为r=a%b，所以a=bq+r，r=a-bq。
a=bq+r，能被b,r整除的，则一定能被a整除，自然也能被a,b整除
r=a-bq，能被a,b整除的，则一定可以被r整除，自然也能被b,r整除
显然gcd(a,b)=gcd(b,r)。
代码很简单：
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int gcd(int a,int b) { int m=a,n=b; if (a < b) { m=b; n=a; } if(m%n==0) return b; else return gcd(n,r); } |
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