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.
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
题意：
求同时满足gcd(x,y,z)==g&&lcm(x,y,z)==L的(x,y,z)组合数.
题解：
1.很直观的：若l%g！=0||g>l时，答案为0；
2.若果满足l%g==0,用唯一分解定理把个g,l分解，用map保存g,l相同因子个数的差
3.对于x,y,z，其对应的质因子pi应该属于[0,s],s为map[pi],pi的个数,而且x,y,z中必须有一个为0，一个为map[pi]，这是为了保证lcm=L,gcd=g;
每个pi对应的种数累乘就是答案了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos(-1.0);
const double E=2.718281828;
typedef long long ll;
const int INF=1000010;
using namespace std;
map<int,int>mp;
void Fenjie(int n,int d)
{
for(int i=2; i*i<=n; i++)
{
if(n%i==0)
{
while(n%i==0)
{
mp[i]+=d;
n/=i;
}
}
if(n==1)
break;
}
if(n>1)
mp[n]+=d;
}
int main()
{
//freopen("in.txt","r",stdin);
int n;
while(cin>>n)
{
while(n--)
{
mp.clear();
int g,l;
cin>>g>>l;
if(g>l||l%g)
{
printf("0\n");
continue;
}
Fenjie(l,1);
Fenjie(g,-1);
ll ans=1;
map<int,int>::iterator it;
for(it=mp.begin();it!=mp.end();it++)
{
if(it->second)
{
int x=it->second;
//ans*=((x+1)*(x+1)*(x+1)-2*x*x*x+(x-1)*(x-1)*(x-1));//容斥原理
ans*=6*x;//组合原理，c(3,1)*c(2,1)*(x-1+1)
}
}
cout<<ans<<endl;
}
}
return 0;
}
My code: