HDU4497GCD and LCM

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 50    Accepted Submission(s): 27

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

1 1 1

A32=6

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <sstream>
#include <map>
#include <bitset>
using namespace std ;
#define zero {0}
#define INF 2000000000
#define eps 1e-6
typedef long long LL;
#define MAX 100000
int prime[MAX]=zero;
int Prime(int a[],int n)
{
    int i,j,k,x,num,*b;
    n++;
    n/=2;
    b=(int *)malloc(sizeof(int)*(n+1)*2);
    a[0]=2;a[1]=3;num=2;
    for(i=1;i<=2*n;i++)
        b[i]=0;
    for(i=3;i<=n;i+=3)
        for(j=0;j<2;j++)
            {
            x=2*(i+j)-1;
            while(b[x]==0)
                {
                a[num++]=x;
                for(k=x;k<=2*n;k+=x)
                    b[k]=1;
                }
            }
    return num;
}
int main()
{
    #ifdef DeBUGs
        freopen("C:\Users\Sky\Desktop\1.in","r",stdin);
    #endif
    int maxn;
    maxn=Prime(prime,MAX);
    int i,j,k;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int gcd,lcm;
        scanf("%d%d",&gcd,&lcm);
        if(gcd==0||lcm==0)
        {
            printf("0
");
            continue;
        }
        if(lcm%gcd)
        {
            printf("0
");
            continue;
        }
        int de=lcm/gcd;
        int num=0;
        int sum=1;
        for(i=0;prime[i]<=de&&i<maxn&&prime[i]>0;i++)
        {
            num=0;
            while(de%prime[i]==0)
            {
                de/=prime[i];
                num++;
            }
            if(!num)
            continue;
            sum*=num*6;
        }
        if(de>1)
        sum*=6;
        printf("%d
",sum);
    }
    
    return 0;
}