【luogu1072】Hankson 的趣味题 [数学]

P1072 Hankson 的趣味题

枚举gcd(x,b0)判断

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<stack>
 7 #include<algorithm>
 8 using namespace std;
 9 #define ll long long
10 #define rg register
11 const int N=2000+5,M=2000000000+5,X=4000;
12 template <class t>void rd(t &x)
13 {
14     x=0;int w=0;char ch=0;
15     while(!isdigit(ch)) w|=ch==‘-‘,ch=getchar();
16     while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
17     x=w?-x:x;
18 }
19
20 int gcd(int a,int b){return b?gcd(b,a%b):a;}
21
22 int main()
23 {
24     //freopen("in.txt","r",stdin);
25     //freopen("nocows.out","w",stdout);
26     int T;rd(T);
27     while(T--)
28     {
29         int a0,a1,b0,b1,ans=0,lim,x;
30         rd(a0),rd(a1),rd(b0),rd(b1);
31         if(b1%b0) {printf("0\n");continue;}
32         for(rg int i=1;i<sqrt(b0);++i)
33         {
34             if(b0%i) continue;
35             x=b1/b0*i;
36             if(gcd(x,b0)==i&&gcd(x,a0)==a1) ++ans;
37             x=b1/b0*(b0/i);
38             if(gcd(x,b0)==b0/i&&gcd(x,a0)==a1) ++ans;
39         }
40         lim=int(sqrt(b0));
41         if(lim*lim==b0&&!(b1%lim))
42         {
43             x=b1/b0*lim;
44             if(gcd(x,b0)==lim&&gcd(x,a0)==a1) ++ans;
45         }
46         printf("%d\n",ans);
47     }
48     return 0;
49 }

原文地址：https://www.cnblogs.com/lxyyyy/p/10879421.html