【BZOJ2301】【HAOI2011】Problem b [莫比乌斯反演]

 1 #include<iostream>  
 2 #include<string>  
 3 #include<algorithm>  
 4 #include<cstdio>  
 5 #include<cstring>  
 6 #include<cstdlib>  
 7 #include<cmath>
 8 using namespace std; 
 9 typedef long long s64;
10   
11 const int ONE = 50005;
12     
13 int T;
14 int Ax,Bx,Ay,By,k;
15 bool isp[ONE];
16 int prime[ONE],p_num;
17 int miu[ONE],sum_miu[ONE];
18 s64 Ans;
19  
20 int get() 
21 {
22         int res=1,Q=1;  char c;
23         while( (c=getchar())<48 || c>57)
24         if(c==‘-‘)Q=-1;
25         if(Q) res=c-48; 
26         while((c=getchar())>=48 && c<=57) 
27         res=res*10+c-48; 
28         return res*Q; 
29 }
30   
31 void Getmiu(int MaxN)
32 {
33         miu[1] = 1;
34         for(int i=2; i<=MaxN; i++)
35         {
36             if(!isp[i])
37                 prime[++p_num] = i, miu[i] = -1;
38             for(int j=1; j<=p_num, i*prime[j]<=MaxN; j++)
39             {
40                 isp[i * prime[j]] = 1;
41                 if(i%prime[j] == 0)
42                 {
43                     miu[i * prime[j]] = 0;
44                     break;
45                 }
46                 miu[i * prime[j]] = -miu[i];
47             }
48             miu[i] += miu[i-1];
49         }
50 }
51  
52 s64 Calc(int n,int m)
53 {
54         if(n > m) swap(n,m);
55          
56         int N = n/k, M = m/k;   Ans = 0;
57         for(int i=1,j=0; i<=N; i=j+1)
58         {
59             j = min(N/(N/i), M/(M/i));
60             Ans += (s64)(N/i) * (M/i) * (miu[j] - miu[i-1]);
61         }
62          
63         return Ans;
64 }
65  
66 void Solve()
67 {
68         Ax=get();   Bx=get();   Ay=get();   By=get();   k=get();
69         printf("%lld\n", Calc(Bx,By) - Calc(Ax-1,By) - Calc(Ay-1,Bx) + Calc(Ax-1,Ay-1));
70 }
71  
72 int main()
73 {
74         Getmiu(ONE-1);
75         T=get();
76         while(T--)
77             Solve();
78 }