文章目录
题意：
给三个数的 l c m lcm lcm和 g c d gcd gcd，求满足条件的三元组组合个数。
思路：
首先
l
c
m
m
o
d
g
c
d
=
=
0
lcm\bmod gcd==0
lcmmodgcd==0是有组合的条件，否则输出0。
现在可知
l
c
m
(
x
′
,
y
′
,
z
′
)
=
l
c
m
(
x
,
y
,
z
)
g
c
d
(
x
,
y
,
z
)
，
g
c
d
(
x
′
,
y
′
,
z
′
)
=
1
lcm(x^{'},y^{'},z^{'})=\frac{lcm(x,y,z)}{gcd(x,y,z)}，gcd(x^{'},y^{'},z^{'})=1
lcm(x′,y′,z′)=gcd(x,y,z)lcm(x,y,z)，gcd(x′,y′,z′)=1，对
a
a
a分解质因子得到
p
1
u
1
p
2
u
2
.
.
.
p
n
u
n
p_1^{u_1}p_2^{u_2}...p_n^{u_n}
p1u1p2u2...pnun，假设
x
′
=
p
1
i
1
p
2
i
2
.
.
.
p
n
j
n
,
y
′
=
p
1
j
1
p
2
j
2
.
.
.
p
n
j
n
,
z
′
=
p
1
k
1
p
2
k
2
.
.
.
p
n
k
n
x^{'}=p_1^{i_1}p_2^{i_2}...p_n^{j_n},y^{'}=p_1^{j_1}p_2^{j_2}...p_n^{j_n},z^{'}=p_1^{k_1}p_2^{k_2}...p_n^{k_n}
x′=p1i1p2i2...pnjn,y′=p1j1p2j2...pnjn,z′=p1k1p2k2...pnkn。那么由于
g
c
d
(
x
′
,
y
′
,
z
′
)
=
1
gcd(x^{'},y^{'},z^{'})=1
gcd(x′,y′,z′)=1，可知
m
i
n
(
i
1
,
j
1
,
k
1
)
=
0
min(i_1,j_1,k_1)=0
min(i1,j1,k1)=0,
m
a
x
(
i
1
,
j
1
,
k
1
)
=
u
1
max(i_1,j_1,k_1)=u_1
max(i1,j1,k1)=u1，所以我们需要找出来一个位置取
0
0
0，一个位置取
u
1
u_1
u1，其他的位置随意就好了。当前位置的答案即为
A
3
2
∗
u
1
=
6
∗
u
1
A_3^2*u_1=6*u_1
A32∗u1=6∗u1，那么
a
n
s
=
∑
6
∗
u
i
ans=\sum6*u_i
ans=∑6∗ui。
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;
//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;
LL g,l;
int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);
int _; scanf("%d",&_);
while(_--)
{
scanf("%lld%lld",&g,&l);
if(l%g!=0) { puts("0"); continue; }
LL x=l/g;
map<int,int>mp;
for(int i=2;i<=x/i;i++)
if(x%i==0)
while(x%i==0) x/=i,mp[i]++;
if(x>1) mp[x]++;
LL ans=1;
for(auto x:mp) ans*=6*x.Y;
printf("%lld\n",ans);
}
return 0;
}
/*
*/