[NOIP2009] Hankson 的趣味题

嘟嘟嘟

 

这题居然暴力就过了……

在[1, sqrt(b1)]中枚举x，判断x | b1，若是成立，再判断lcm(x, b0) == b1 && gcd(x, a0) == a1，并计数。

复杂度O(n * sqrt(b1) * logb1)。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<stack>
 9 #include<queue>
10 #include<vector>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(' ')
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 0x3f3f3f3f;
19 const db eps = 1e-8;
20 //const int maxn = ;
21 inline ll read()
22 {
23   ll ans = 0;
24   char ch = getchar(), las = ' ';
25   while(!isdigit(ch)) las = ch, ch = getchar();
26   while(isdigit(ch)) ans = ans * 10 + ch - '0', ch = getchar();
27   if(las == '-') ans = -ans;
28   return ans;
29 }
30 inline void write(ll x)
31 {
32   if(x < 0) putchar('-'), x = -x;
33   if(x >= 10) write(x / 10);
34   putchar(x % 10 + '0');
35 }
36 
37 int n;
38 ll a0, a1, b0, b1;
39 
40 ll gcd(ll a, ll b)
41 {
42   return b ? gcd(b, a % b) : a;
43 }
44 
45 int main()
46 {
47   n = read();
48   for(int i = 1; i <= n; ++i)
49     {
50       a0 = read(); a1 = read(); b0 = read(); b1 = read();
51       int cnt = 0;
52       for(int j = 1; j * j <= b1; ++j)
53     {
54       if(!(b1 % j))
55         {
56           ll Lcm1 = j / gcd(j, b0) * b0, Gcd1 = gcd(j, a0);
57           ll Lcm2 = b1 / j / gcd(b1 / j, b0) * b0, Gcd2 = gcd(b1 / j, a0);
58           if(j * j == b1)
59         {
60           if(Lcm1 == b1 && Gcd1 == a1) cnt++;
61         }
62           else
63         {
64           if(Lcm1 == b1 && Gcd1 == a1) cnt++;
65           if(Lcm2 == b1 && Gcd2 == a1) cnt++;
66         }
67       }
68     }
69       write(cnt); enter; 
70     }
71   return 0;
72 }

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