hdu4497-GCD and LCM-(欧拉筛+唯一分解定理+组合数)

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3409    Accepted Submission(s):
1503

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

 

翻译：给出g和l，求满足gcd(x,y,z)=g,lcm(x,y,z)=l的(x,y,z)组合的数量，(1, 2, 3)和(1, 3, 2)算作不同组合

分析：

g是因数，l是倍数，显然可以得到x%g=y%g=z%g=0=l%x=l%y=l%z;

进一步推出l%g=0;若不符合，无解。

 

根据唯一分解定理

将x，y，z分解

x=p1^a1*p2^a2……ps^as; 
y=p1^b1*p2^b2……ps^bs; 
z=p1^c1*p2^c2……ps^cs; 

g是三个数的最大公约数，则g满足g=p1^{min(a1,b1,c1)}……ps^{min(as,bs,cs)}=p1^e1……ps^es 

l是三个数的最小公倍数，则l满足l=p1^{max(a1,b1,c1)}……ps^{max(as,bs,cs)}=p1^h1……ps^hs

ei=min(ai,bi,ci) hi=max(ai,bi,bi)

x，y，z三个数分解后，对于每一个质因子，必然有一个指数是ei，一个指数是hi，先将这两个数固定下来

另一个数可以取ei到hi之间的数，包括ei和hi，设为ti

(1)当ti=ei或者ti=hi时，(ei,ei,hi)只有三种组合方式，(ei,hi,hi)也只有三种组合方式

(2)当ti≠ei并且ti≠hi时，(ei,ti,hi)有6种组合方式

(3)当ei=ti=hi时，只有1种组合方式

当ei≠hi时，则每个质因子全部的组合方式有6*(hi-ei)种，组合数用乘法计算结果，当hi=ei，不累乘0

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<string>
#include<vector>
#include<iostream>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;
#define ll long long
const int maxx=1e6+;
int prime[maxx];
int vis[maxx];
ll g,l;
int cnt;
void init()
{
    memset(vis,true,sizeof(vis));
    vis[]=vis[]=false;
    cnt=;
    for(int i=;i<=maxx;i++)
    {
        if(vis[i])
            prime[cnt++]=i;
        for(int j=;j<cnt && prime[j]*i<=maxx;j++)
        {
            vis[ i*prime[j] ]=false;
            if(i%prime[j]==) break;
        }
    }
}

int main()///hdu4497
{
    init();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll ans=;
        scanf("%lld%lld",&g,&l);
        if(l%g)
            printf("0\n");
        else
        {
            for(int i=;i<cnt;i++)
            {
                int e=,h=;
                while(g%prime[i]==)
                {
                    e++;
                    g=g/prime[i];
                }
                while(l%prime[i]==)
                {
                    h++;
                    l=l/prime[i];
                }
                if(h-e)
                {
                    ans=ans*(h-e)*;
                }
            }
            if(g==l)///32位范围内大素数因子只能有一个，l能被g整除证明这个大素数相同，不累乘
                ;
            else if(l>)///如果g=1，l=大素数，再乘一次6
                ans=ans*;
            printf("%lld\n",ans);
        }
    }
    return ;
}