GreatestCommonDivisor 最大公约数

最大公约数/最小公倍数

最大公约数gcdgcd是最大的能同时被a,ba, b整除的正整数。最小公倍数lcmlcm是最小的能同时整除a,ba, b的正整数。即满足下列等式:

amodgcd=0bmodgcd=0lcmmoda=0lcmmodb=0\begin{matrix} a \bmod gcd = 0 \\ b \bmod gcd = 0 \\ lcm \bmod a = 0 \\ lcm \bmod b = 0 \end{matrix}

00和另一个正整数bb的最大公约数为bb,因为00除以任何正整数都可以整除。

对于a=36a = 36,能被aa整除的正整数为divisorsa=[1,2,3,4,6,9,12,18,36]divisors_{a} = [1, 2, 3, 4, 6, 9, 12, 18, 36],能整除aa的正整数为multiplesa=[36,72,108]multiples_{a} = [36, 72, 108 \dots]。对于b=66b = 66,能被bb整除的正整数为divisorsb=[1,2,3,6,11,22,33,66]divisors_{b} = [1, 2, 3, 6, 11, 22, 33, 66],能整除bb的正整数为multiplesb=[66,132,198]multiples_{b} = [66, 132, 198 \dots]。那么36,6636, 66的最大公约数为66,最小公倍数为396396

a=36,b=66a = 36, b = 66分解为素数的乘积,可得36=2×2×3×3,66=2×3×1136 = 2 \times 2 \times 3 \times 3, 66 = 2 \times 3 \times 11,其中每个素数都称为分解因子。可以把aabb的分解因子看作两个集合factora=[2,2,3,3],factorb=[2,3,11]factor_{a} = [2, 2, 3, 3], factor_{b} = [2, 3, 11]。则a,ba, b的最大公约数为两个正整数分解因子的最大交集,最小公倍数为最小并集

gcd(a,b)=factorafactorblcm(a,b)=factorafactorb\begin{matrix} gcd(a, b) = factor_{a} \bigcap factor_{b} \\ lcm(a, b) = factor_{a} \bigcup factor_{b} \end{matrix}

分解因子是一个NP完全问题。

最大公约数、最小公倍数具有以下性质:

gcd(a,b)×lcm(a,b)=a×bgcd(a,lcm(b,c))=lcm(gcd(a,b),gcd(a,c))lcm(a,gcd(b,c))=gcd(lcm(a,b),lcm(a,c))\begin{matrix} gcd(a,b) \times lcm(a,b) & = & a \times b \\ gcd(a, lcm(b, c)) & = & lcm(gcd(a, b), gcd(a, c)) \\ lcm(a, gcd(b, c)) & = & gcd(lcm(a, b), lcm(a, c)) \end{matrix}

比如a=24,b=20,c=30a = 24, b = 20, c = 30,满足以下等式:

gcd(a,b)×lcm(a,b)=4×120=24×20=480gcd(a,lcm(b,c))=gcd(24,60)=12=lcm(gcd(a,b),gcd(a,c))=lcm(4,6)lcm(a,gcd(b,c))=lcm(24,10)=120=gcd(lcm(a,b),lcm(a,c))=gcd(120,120)\begin{matrix} gcd(a, b) \times lcm(a, b) & = & 4 \times 120 & = & 24 \times 20 & = & 480 \\ gcd(a, lcm(b, c)) & = & gcd(24, 60) & = & 12 & = & lcm(gcd(a, b), gcd(a, c)) & = & lcm(4, 6) \\ lcm(a, gcd(b, c)) & = & lcm(24, 10) & = & 120 & = & gcd(lcm(a, b), lcm(a, c)) & = & gcd(120, 120) \end{matrix}

可知a,ba, b的最小公倍数与最大公倍数的关系为lcm=a×b÷gcdlcm = a \times b \div gcd

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