BZOJ2820: YY的GCD

Posted 2020-10-23 w-h-h

莫比乌斯反演

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<vector>
 6 #include<cmath>
 7 #include<queue>
 8 #define MAXN 10000000+10
 9 #define INF 0x7f7f7f7f
10 #define LINF 0x7f7f7f7f7f7f7f7f
11 #define ll long long
12 #define pb push_back
13 #define ft first
14 #define sc second
15 #define mp make_pair
16 #define pil pair<int,ll>
17 #define pll pair<ll,ll>
18 using namespace std;
19 int f[MAXN],mu[MAXN],b[MAXN];
20 vector<int> p;
21 void getmu(){
22     mu[1]=1;
23     for(int i=2;i<MAXN;i++){
24         if(!b[i]){
25             p.pb(i);
26             mu[i]=-1;
27         }
28         for(int j=0;j<p.size()&&p[j]*i<MAXN;j++){
29             b[p[j]*i]=1;
30             if(i%p[j]==0){
31                 mu[p[j]*i]=0;
32                 break;
33             }
34             mu[p[j]*i]=-mu[i];
35         }
36     }
37     for(int i=0;i<p.size();i++){
38         for(int j=1;j*p[i]<MAXN;j++){
39             f[j*p[i]]+=mu[j];
40         }
41     }
42     for(int i=1;i<MAXN;i++){
43         f[i]+=f[i-1];
44     }
45 }
46 int n,m;
47 void solve(){
48     scanf("%d%d",&n,&m);
49     if(n>m)swap(n,m);
50     int lst;
51     ll ans=0LL;
52     for(int i=1;i<=n;i=lst+1){
53         lst=min(n/(n/i),m/(m/i));
54         ans+=1LL*(f[lst]-f[i-1])*(n/i)*(m/i);
55     }
56     printf("%lld\n",ans);
57 }
58 int main()
59 {
60     getmu();
61     int T;
62     scanf("%d",&T);
63     while(T--){
64         solve();
65     }
66     return 0;
67 }