HDOJ 4497 GCD and LCM

组合数学：GCD与LCM问题求解

最新推荐文章于 2019-12-05 21:18:40 发布

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最新推荐文章于 2019-12-05 21:18:40 发布

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文章标签： java

原文链接：http://www.cnblogs.com/CKboss/p/3373031.html

本文探讨了在给定整数G和L的情况下，如何找出满足特定条件(x, y, z)的解决方案数量。通过计算最大公约数(GCD)和最小公倍数(LCM)，读者将了解如何解决此类数学问题。

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

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

6 72

7 33

Sample Output

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

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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 }