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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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