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：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 1111111;
int nu[maxn];
int isprim[maxn];
int prim[maxn];
int main()
{
    int t;
    int g,l;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&g,&l);
        if(l%g!=0)
        {
            printf("0
");
            continue;
        }
        memset(nu,0,sizeof(nu));
        int cnt=0;
        
        int m=l/g;
        int sqm=sqrt(m+1);
        for(int i=2;i<=sqm;++i)
        {
            if(m%i==0)
            {
                ++cnt;
                while(m%i==0)
                {
                    nu[cnt]++;
                    m/=i;
                }
            }
            
        }ll sum=1;
        if(m!=1)
            sum=6;
        
        for(int i=1;i<=sqm;++i)
        {
            if(nu[i]!=0)
            sum*=nu[i]*6;
        }
        printf("%lld
",sum);
    }
    return 0;
}