【莫比乌斯反演】BZOJ2154 Crash的数字表格

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #define ll long long
 5 using namespace std;
 6 const int maxn=1e7+5,mod=20101009;
 7 
 8 bool flag[maxn];
 9 int prime[maxn],mu[maxn],cnt;
10 int sum[maxn],s[maxn];
11 int n,m;
12 
13 void getmu(){
14     mu[1]=1;
15     for(int i=2;i<=n;i++){
16         if(!flag[i]){
17             mu[i]=-1;
18             prime[++cnt]=i;
19         }
20         for(int j=1;i*prime[j]<=n&&j<=cnt;j++){
21             flag[i*prime[j]]=1;
22             if(i%prime[j]==0){
23                 mu[i*prime[j]]=0;
24                 break;
25             }
26             mu[i*prime[j]]=-mu[i];
27         }
28     }
29     for(int i=1;i<=n;i++)
30         sum[i]=(sum[i-1]+(ll)i*i*mu[i]%mod)%mod;
31 }
32 
33 ll F(int x,int y){
34     return (ll)((ll)x*(x+1)/2%mod)*((ll)y*(y+1)/2%mod)%mod;
35 }
36 
37 ll f(int x,int y){
38     ll ret=0;
39     int p;
40     for(int i=1;i<=x;i=p+1){
41         p=min(x/(x/i),y/(y/i));
42         ret=(ret+(ll)(sum[p]-sum[i-1])*F(x/i,y/i))%mod;
43     }
44     return ret;
45 }
46 
47 int main(){
48     scanf("%d%d",&n,&m);
49     if(n>m) swap(n,m);
50     for(int i=1;i<=n;i++) s[i]=(s[i-1]+i)%mod;
51     getmu();
52     
53     int pos;
54     ll ans=0;
55     for(int i=1;i<=n;i=pos+1){
56         pos=min(n/(n/i),m/(m/i));
57         ans=(ans+(ll)(s[pos]-s[i-1])*f(n/i,m/i))%mod;
58     }
59     printf("%lld\n",(ans+mod)%mod);
60     return 0;
61 }