#最大公约数
之前,我们学过约数,这次,我们需要找到2个数a,ba, ba,b的最大的公约数,我们设d=gcd(a,b)d = gcd (a,b)d=gcd(a,b)为a,ba,ba,b的最大公约数
所以有d∣ad | ad∣a,d∣bd | bd∣b,设a=kb+ra = kb + ra=kb+r,其中k为整数
所以r=a−kbr = a - kbr=a−kb,根据整除的性质,可得d∣rd | rd∣r
故d也是r的公因数,若gcd(b,r)=cgcd (b, r) = cgcd(b,r)=c,则d<=cd <= cd<=c
又有c∣rc | rc∣r,c∣bc | bc∣b,a=kb+ra = kb + ra=kb+r,所以c∣ac | ac∣a
所以c是a、b的公约数,即c<=dc <= dc<=d,综上所以c=dc = dc=d
所以gcd(a,b)=gcd(b,a mod b),同理也可证=gcd(b mod a,a)gcd (a,b) = gcd (b,a \:mod \: b) ,同理也可证= gcd (b \: mod \: a, a)gcd(a,b)=gcd(b,amodb),同理也可证=gcd(bmoda,a)
###code
#include <iostream>
using namespace std;
int gcd1 (int a, int b) {
return b ? gcd1(b, a % b) : a;
}
int gcd2 (int a,int b) {
return a ? gcd2(b % a, a) : b;
}
int main () {
int m,n;
cin >> m >> n;
cout << gcd1 (m,n) << " " << gcd2 (m,n);
}
#最小公倍数
设a,ba, ba,b的最小公倍数为LLL,最大公因数为DDD
所以有L∣aL|aL∣a,L∣bL|bL∣b
且有a=Dxa = Dxa=Dx,b=Dyb = Dyb=Dy(显然有x,yx,yx,y互质)
故满足L∣D, L∣x, L∣yL|D,\:L|x,\:L|yL∣D,L∣x,L∣y,所以满足条件的最小的L=D×x×y=a÷D×bL=D \times x \times y = a \div D \times bL=D×x×y=a÷D×b
其实这里可以使用唯一分解定理证明
设
a=p1a1×p2a2×...×pnana = p_{1}^{a_1} \times p_{2}^{a_2} \times ... \times p_{n}^{a_n}a=p1a1×p2a2×...×pnan
b=p1b1×p2b2×...×pnbnb = p_{1}^{b_1} \times p_{2}^{b_2} \times ... \times p_{n}^{b_n}b=p1b1×p2b2×...×pnbn
则明显D=gcd(a,b)=p1min(a1,b1)×p2min(a2,b2)×...×pnmin(an,bn)D = gcd(a,b) = p_{1}^{min(a_1,b_1)} \times p_{2}^{min(a_2,b_2)} \times ... \times p_{n}^{min(a_n,b_n)}D=gcd(a,b)=p1min(a1,b1)×p2min(a2,b2)×...×pnmin(an,bn)
而L=lcm(a,b)=p1max(a1,b1)×p2max(a2,b2)×...×pnmax(an,bn)L = lcm(a,b) = p_{1}^{max(a_1,b_1)} \times p_{2}^{max(a_2,b_2)} \times ... \times p_{n}^{max(a_n,b_n)}L=lcm(a,b)=p1max(a1,b1)×p2max(a2,b2)×...×pnmax(an,bn)
可得L=a÷D×bL=a \div D \times bL=a÷D×b