给出方程组：
gcd(x,a0)=a1
lcm(x,b0)=b1
在保证a1是a0的因数且b0是b1的因数的情况下，求解满足条件的x有多少个
考虑到x既是a1的倍数又是b1的因数，可以sqrt(b1)枚举因数然后检查是否满足条件
注意for循环里得先判断是不是因数，否则会爆T
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int gcd(int a, int b) { if(a == 0) return b; if(b == 0) return a; if(!(a & 1) && !(b & 1)) return gcd(a >> 1, b >> 1) << 1; else if(!(b & 1)) return gcd(a, b >> 1); else if(!(a & 1)) return gcd(a >> 1, b); else return gcd(abs(a - b), min(a, b)); } int lcm(int x,int y){return x*y/gcd(x,y);} int t; int a0,a1,b0,b1; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); t=read(); while(t--){ a0=read(); a1=read(); b0=read(); b1=read(); int p=a0/a1,q=b1/b0,ans=0; for(int i=1;i*i<=b1;++i){ if(b1%i==0){ if(i%a1==0&&gcd(i/a1,p)==1&&gcd(q,b1/i)==1) ++ans; if(b1/i==i) continue; if(b1/i%a1==0&&gcd(b1/i/a1,p)==1&&gcd(q,b1/(b1/i))==1) ++ans; } } write(ans),prn(); } return 0; }