HDOJ 4497 GCD and LCM

组合数学

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 451    Accepted Submission(s): 216


[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
 
 

[align=left]Recommend[/align]
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1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4
5 using namespace std;
6
7 int main()
8 {
9     int t,L,G;
10     scanf("%d",&t);
11     while(t--)
12     {
13         scanf("%d%d",&G,&L);
14         if(L%G!=0)
15         {
16             puts("0");
17             continue;
18         }
19         int sk=L/G;
20         int pp=2,ans=1,cnt;
21         while(sk!=1)
22         {
23             cnt=0;
24             while(sk%pp==0)
25             {
26                 cnt++;
27                 sk/=pp;
28             }
29             pp++;
30             if(cnt!=0)
31             {
32                 ans*=cnt*6;
33             }
34         }
35         printf("%d\n",ans);
36     }
37     return 0;
38 }