【反演复习计划】【COGS2432】爱蜜莉雅的施法

 1 #include<bits/stdc++.h>
 2 #define N 10000010
 3 using namespace std;
 4 typedef long long ll;
 5 int n,m,prime[N],cnt,vis[N];
 6 ll f[N];
 7 void calcpre(){
 8     f[1]=1;cnt=0;memset(vis,1,sizeof(vis));
 9     for(int i=2;i<=N;i++){
10         if(vis[i]){prime[++cnt]=i;f[i]=i-2;}
11         for(int j=1;j<=cnt;j++){
12             int t=i*prime[j];if(t>N)break;
13             vis[t]=0;int p=prime[j];
14             if (i%p)f[t]=f[i]*f[p];
15             else{
16                 f[t]=(i/p)%p?f[i/p]*(p-1)*(p-1):f[i]*p;
17                 break;
18             }
19         }
20     }
21     for(int i=2;i<=N;i++)f[i]+=f[i-1];
22 }
23 inline int read(){
24     int f=1,x=0;char ch;
25     do{ch=getchar();if(ch==‘-‘)f=-1;}while(ch<‘0‘||ch>‘9‘);
26     do{x=x*10+ch-‘0‘;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘);
27     return f*x;
28 }
29 int main(){
30     freopen("aimiliyausemagic.in","r",stdin);
31     freopen("aimiliyausemagic.out","w",stdout);
32     calcpre();int T=read();
33     while(T--){
34         n=read();m=read();
35         if(n>m)swap(n,m);ll ans=0;
36         for(int i=1,j=1;i<=n;i=j+1){
37             j=min(n/(n/i),m/(m/i));
38             ans+=(f[j]-f[i-1])*(n/i)*(m/i);
39         }
40         printf("%lld\n",ans);
41     }
42     return 0;
43 }