hdu4497 GCD and LCM
GCD and LCM
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 78 Accepted Submission(s): 43
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.
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.
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
Recommend
liuyiding
很明显，m/n!=0的话，就直接输出0就可以了！否刚，直接分解质因数m/n,找到，每个质因子的个数，这样，我们，就可以得出每个质因数为a1^k1,那是题目就是要把这k1个a1分到三个数中，那么排列组合就是k1*A（3，2）,也就是，6*k1,种，直接算出来就行了！
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
using namespace std;
#define MAXN 100000
int num[MAXN];
int main()
{
int tcase,n,m,i,ans,sum,tempm;
scanf("%d",&tcase);
while(tcase--)
{
scanf("%d%d",&n,&m);
memset(num,0,sizeof(num));
if(m%n!=0)
{
printf("0\n");
continue;
}
m=m/n;tempm=sqrt(m)+1;
for(i=2,ans=0;i<=tempm;i++)
{
if(m%i==0)
{
while(m%i==0)
{
num[ans]++;m/=i;
}
ans++;
}
}
if(m!=1)
num[ans++]=1;
for(sum=1,i=0;i<ans;i++)
sum*=6*num[i];
printf("%d\n",sum);
}
return 0;
}