【反演复习计划】【bzoj4407】于神之怒加强版

#include<bits/stdc++.h>
#define N 5000010
#define yql 1000000007
using namespace std;
typedef long long ll;
int T,k,n[],m[],maxn,vis[N],prime[N];
int mu[N],cnt=,fac[N];
ll f[N],s[N];
inline int read(){
    int f=,x=;char ch;
    do{ch=getchar();if(ch=='-')f=-;}while(ch<''||ch>'');
    do{x=x*+ch-'';ch=getchar();}while(ch>=''&&ch<='');
    return f*x;
}
inline ll fpow(ll x,ll p){
    x%=yql;ll ans=;
    for(;p;p>>=,x=x*x%yql)if(p&)ans*=x,ans%=yql;
    return (ans+yql)%yql;
}
void calcmu(){
    cnt=;mu[]=;s[]=;memset(vis,,sizeof(vis));
    for(int i=;i<=maxn;i++){
        if(vis[i]){prime[++cnt]=i;fac[i]=i;mu[i]=-;f[i]=fpow(i,k);s[i]=f[i]-;}
        for(int j=;j<=cnt;j++){
            int p=prime[j],t=i*p;if(t>maxn)break;
            vis[t]=;
            if(i%p==){
                mu[t]=;fac[t]=fac[i]*p;
                s[t]=1LL*s[i]*f[prime[j]]%yql;
                break;
            }
            mu[t]=-mu[i];s[t]=1LL*s[i]*s[p]%yql;
        }
    }
    for(int i=;i<=maxn;i++)mu[i]+=mu[i-],s[i]+=s[i-],s[i]%=yql;
}
int main(){
    T=read();k=read();
    for(int i=;i<=T;i++){
        n[i]=read(),m[i]=read();
        maxn=max(maxn,max(n[i],m[i]));
    }
    calcmu();
    for(int i=;i<=T;i++){
        int nn=n[i],mm=m[i];if(nn>mm)swap(nn,mm);
        ll ans=;
        for(int i=,j=;i<=nn;i=j+){
            j=min(mm/(mm/i),nn/(nn/i));
            ans+=(s[j]-s[i-]+yql)%yql*(nn/i)%yql*(mm/i)%yql;ans%=yql;
        }
        printf("%lld\n",ans);
    }
}

试了下maxn操作，结果还是BZOJ倒数TAT

感觉式子化得没问题啊？