hdu 4497 GCD and LCM （非原创）

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2977    Accepted Submission(s): 1302

Problem Description

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.

 

Input

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.

 

Output

For each test case, print one line with the number of solutions satisfying the conditions above.

 

Sample Input

2
6 72
7 33

 

Sample Output

72
0

 

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

 

Recommend

liuyiding

 

 

这是一道好题。

 

我是看题解才会的：https://www.cnblogs.com/fightfor/p/3960212.html

思路

将满足条件的一组x,z,y都除以G,得到x‘，y',z',满足条件gcd(x',y',x') = 1,同时lcm(x',y',x') = G/L.
特判，当G%L ！= 0 时，无解。
然后素数分解G/L，假设G/L = p1^t1 * p2^t2 *````* pn^tn。
满足上面条件的x,y,z一定为这样的形式。
x' = p1^i1 * p2^i2 *```* pn^in.
y' = p1^j1 * p2^j2 * ```*pn^jn.
z' = p1^k1 * p2^k2 * ```*pn^kn.
为了满足上面的条件，对于p1,一定有max(i1,j1,k1) = t1.min(i1,j1,k1) =0;则当选定第一个数为0，第二个数为t1时，第三个数可以为0-t1，又由于有顺序的，只有(0,t1,t1) 和(0,t1,0)这两种情形根据顺序只能产生三种结果，其他的由于三个数都不一样，一定能产生6种，所以最后产生了6*（t1-1）+3*2 = 6*t1种，根据乘法原理以及关于素数分解的唯一性，反过来，素数组合必然也是唯一的数，一共有6*t1 * 6*t2 *`````*6*tn种选法。

附ac代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <cmath>
 6 using namespace std;
 7 typedef long long ll;
 8 const int maxn = 1111111;
 9 int nu[maxn];
10 int isprim[maxn];
11 int prim[maxn];
12 int main()
13 {
14     int t;
15     int g,l;
16     scanf("%d",&t);
17     while(t--)
18     {
19         scanf("%d%d",&g,&l);
20         if(l%g!=0)
21         {
22             printf("0\n");
23             continue;
24         }
25         memset(nu,0,sizeof(nu));
26         int cnt=0;
27
28         int m=l/g;
29         int sqm=sqrt(m+1);
30         for(int i=2;i<=sqm;++i)
31         {
32             if(m%i==0)
33             {
34                 ++cnt;
35                 while(m%i==0)
36                 {
37                     nu[cnt]++;
38                     m/=i;
39                 }
40             }
41
42         }ll sum=1;
43         if(m!=1)
44             sum=6;
45
46         for(int i=1;i<=sqm;++i)
47         {
48             if(nu[i]!=0)
49             sum*=nu[i]*6;
50         }
51         printf("%lld\n",sum);
52     }
53     return 0;
54 }