[BZOJ 2154]Crash的数字表格：莫比乌斯反演

点击这里查看原题

太强了……还是看PoPoQQQ的题解吧

于是进行两次分块，求F(x,y)为sqrt(n),求ans也为sqrt(n)，总复杂度O(n)

/*User:SmallLanguage:C++Problem No.:2154*/#include<bits/stdc++.h>#define ll long long#define inf 999999999using namespace std;const int M=1e7+5;const ll mod=20101009;int n,m,mu[M],prime[M],cnt;ll ans,sum[M];bool not_prime[M];ll gets(int n,int m){    n=(ll)n*(n+1)/2%mod;    m=(ll)m*(m+1)/2%mod;    return (ll)n*m%mod;}ll getf(int n,int m){    ll res=0;    for(int i=1,r;i<=n;i=r+1){        r=min(n/(n/i),m/(m/i));        res=(res+(sum[r]-sum[i-1]+mod)%mod*gets(n/i,m/i)%mod)%mod;    }    return res;}int main(){    freopen("data.in","r",stdin);//    scanf("%d%d",&n,&m);    if(n>m) swap(n,m);    mu[1]=1;    for(int i=2;i<=n;i++){        if(!not_prime[i]){            prime[++cnt]=i;            mu[i]=-1;        }        for(int j=1;j<=cnt&&i*prime[j]<=n;j++){            not_prime[i*prime[j]]=1;            if(i%prime[j]==0){                mu[i*prime[j]]=0;                break;            }            mu[i*prime[j]]=-mu[i];        }    }    for(int i=1;i<=n;i++) sum[i]=(sum[i-1]+(ll)i%mod*i%mod*mu[i]%mod)%mod;    for(int i=1,r;i<=n;i=r+1){        r=min(n/(n/i),m/(m/i));        ans=(ans+(ll)(i+r)*(r-i+1)/2%mod*getf(n/i,m/i)%mod)%mod;    }    printf("%lld\n",ans);    return 0;}