【BZOJ2820】YY的GCD [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 = 10000005;
12      
13 int T;
14 int n,m;
15 bool isp[ONE];
16 int prime[700005],p_num;
17 int miu[ONE],sum[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         }
49         for(int j=1; j<=p_num; j++)
50             for(int i=1; i*prime[j]<=MaxN; i++)
51                 sum[i * prime[j]] += miu[i];
52         for(int i=1; i<=MaxN;i++)
53             sum[i] += sum[i-1]; 
54 }
55  
56 void Solve()
57 {
58         n=get();    m=get();
59         if(n > m) swap(n,m);
60         Ans = 0;
61         for(int i=1, j=0; i<=n; i=j+1)
62         {
63             j = min(n/(n/i), m/(m/i));
64             Ans += (s64) (n/i) * (m/i) * (sum[j] - sum[i-1]);
65         }
66         printf("%lld\n",Ans);
67 }
68  
69 int main()
70 {
71         Getmiu(ONE-1);
72         T=get();
73         while(T--)
74             Solve();
75 }