hdu 4497(最大公约数和最小公倍数)

这个题目有个数学知识是以前不知道的。。

Problem E : GCD and LCM
From:HDU, 4497
[align=left]Problem Description[/align]
Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L?

Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z.

Note 2, (1, 2, 3) and (1, 3, 2) are two different solutions.

[align=left]Input[/align]
First line comes an integer T (T <= 12), telling the number of test cases.

The next T lines, each contains two positive 32-bit signed integers, G and L.

It’s guaranteed that each answer will fit in a 32-bit signed integer.

[align=left]Output[/align]
For each test case, print one line with the number of solutions satisfying the conditions above.

[align=left]Sample Input[/align]

2
6 72
7 33

[align=left]Sample Output[/align]

72
0

[align=left]Source[/align]
2013
ACM-ICPC吉林通化全国邀请赛――题目重现

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liuyiding

对于x,y,z都可以写成x = p1^a1+p2^a2+p3^a3....pn^an;y = p1^b1+p2^b2+p3^b3....pn^bn;z=p1^c1+p2^c2+p3^c3....pn^cn;(p1,p2,..pn都是质数也就是传说中的分解质因数)

x,y,z的最大公约数可以写成gcd(x,y,z) = p1^min(a1,b1,c1)*p2^min(a2,b2,c2)......pn^min(an,bn,cn);

最小公倍数则是lcm(x,y,z) = p1^max(a1,b1,c1)*p2^max(a2,b2,c2)....*pn^max(an,bn,cn)

对于gcd(x,y,z) = g,lcm(x,y,z) = l的组合数等gcd(x,y,z) = 1,lcm(x,y,z) = l/g;

设l/g = k;

k = p1^d1+p2^d2+.....+pn^dn;

这样对于p1来说a1,b1,c1中一定有一个为0，因为gcd(x,y,z) = 1,同理也一定有一个为d1;

分为三种情况分别是2个0一个d1,2个d1一个0,1个d1一个0另外一个是[1，(d1-1)]中的一个数，则情况数便是3+3+(d1-1)*6 = 6*d1;

最终答案便是ans = 1;

for(int i=1;i<=n;i++)

ans*=di*6;