HDU 4497 GCD and LCM(拆素數+組合)

畫船聽雨發表於2014-03-22
GC
解體的思路：

lcm(x,y,z)=k;

gcd(x,y,z)=t;

若：x=a*t;   y=b*t;       z=c*t;

則lcm(a,b,c)=k/t;

若k/t=2^A;

則a,b,c中至少有一個數為2^A,至少有一個數是2^0,另外一個數為2^(0~A);共6*A種情況。

則，若k/t=2^A*3^B*5^C;

a,b,c的情況數為：(6*A)*(6*B)*(6*C);

 PS：從鵬哥那裡偷來的，他說的比較清楚。

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 690    Accepted Submission(s): 319


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





 


#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 1000100
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

using namespace std;

const int maxn = 5000010;

int f[maxn];
int p[maxn];
int t;
void prime()
{
    t = 0;
    int n = maxn/2;
    memset(f, 0 , sizeof(f));
    for(int i = 4; i <= n; i += 2)
        f[i] = 1;
    for(int i = 3; i <= n; i +=  2)
    {
        for(int j = 2*i; j <= n; j += i)
            f[j] = 1;
    }
    for(int i = 2; i <= n; i++)
        if(!f[i])
        p[t++] = i;
}
int num[100010];
int main()
{
    prime();
    int T;
    cin >>T;
    while(T--)
    {
        int n, k;
        cin >>n>>k;
        int site = 0;
        memset(num, 0, sizeof(num));
        if(n == k)
            cout<<"1"<<endl;
        else if(k%n)
            cout<<"0"<<endl;
        else
        {
            int m = k/n;
            for(int i = 0; i < t; i++)
            {
                int flag = 0;
                while(1)
                {
                    if(m%p[i] == 0)
                    {
                        m/=p[i];
                        flag = 1;
                        num[site]++;
                    }
                    else
                    {
                        if(flag)
                        site++;
                        break;
                    }
                }
            }
            if(m != 1)
                num[site++] = 1;
            LL sum = 1;
            for(int i = 0; i < site; i++)
                sum *= num[i]*6;
            cout<<sum<<endl;
        }
    }
}




            




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