DZY Loves Math IV

Description

給定n,m，求∑ni=1∑mj=1φ(i∗j) mod 1e9+7
n<=100,000,m<=1,000,000,000

定義sum(n,m)表示∑mi=1φ(i∗n)

令z(n)表示和n含有的質因子相同且指數都爲1的數
則有sum(n,m)=n/z(n)*sum(z(n),m)
那麼現在只需考慮n的質因子指數都爲1的情況。
若素數p|n則有sum(n,m)=(p-1)*sum(n/p,m)+sum(n,m/p)
然後遞歸邊界用杜教篩計算就好了。

代碼

#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define ll long long
using namespace std;
const int maxn=1000000;const int mo=1000000000+7;const int m1=1000007;
int c[maxn/2],fi[maxn+1],z[maxn+1],z1[maxn+1],i,j,n,m,ans;
int h1[m1],h2[m1],w1[m1],w2[m1][2],s2[maxn];
bool bz[maxn+1];
int ha(int n){
    int y=n%m1;
    while (w1[y]!=0&&w1[y]!=n) y=(y+1)%m1;
    return y;
}
int ss(int n){
    if (n<=maxn) return fi[n];
    int j=ha(n);if (w1[j]!=0) return h1[j];w1[j]=n;
    int s=(ll)n*(n+1)/2%mo,i=1;
    while (i<n) {
        i++;int l=n/(n/i)-i;
        s=(s-(ll)(l+1)*ss(n/i)%mo)%mo,i+=l;
    } 
    return h1[j]=s;
}
int sum(int n,int m){
    if (m==0) return 0;
    if (n==1) return ss(m);
    return ((ll)(z[n]-1)*sum(n/z[n],m)%mo+sum(n,m/z[n]))%mo;
}
int main(){
    z1[1]=fi[1]=1;
    fo(i,2,maxn) {
        if (!bz[i]) c[++c[0]]=i,z1[i]=z[i]=i,fi[i]=i-1;
        fo(j,1,c[0]) {
            if (i>maxn/c[j]) break;
            bz[i*c[j]]=1;
            z[i*c[j]]=c[j];
            if (i%c[j]==0){
                z1[i*c[j]]=z1[i];
                fi[i*c[j]]=fi[i]*c[j];
                break;
            }z1[i*c[j]]=z1[i]*c[j];
            fi[i*c[j]]=fi[i]*fi[c[j]];
        }
    }
    fo(i,2,maxn) (fi[i]+=fi[i-1])%=mo;
    scanf("%d%d",&n,&m);
    fo(i,1,n) {
        int q=z1[i];if (s2[q]==0) s2[q]=sum(q,m);
        ans=(ans+(ll)(i/q)*s2[q]%mo)%mo;
    }
    printf("%d\n",(ans%mo+mo)%mo);
}