51nod 1238 最小公倍数之和 V3_综合

我们首先计算

f (n) = = = \sum i = 1 n lcm (i, n) \sum d | n \sum i = 1 n d n i d d [gcd (i, n d) = 1] n \sum d | n d ? φ ( d ) + e ( d ) 2

$\begin{align} f(n)= &\sum_{i=1}^n\text{lcm}(i,n) \\ = & \sum_{d|n}\sum_{i=1}^{\frac n d}\frac{nid}d[\gcd(i,\frac n d)=1] \\ = & n\sum_{d|n}\frac{d\cdot\varphi(d)+e(d)}2 \end{align}$
于是

= = \sum i = 1 n f (i) 1 2 \sum d = 1 n d ? φ (d) + e (d) \sum i = 1 ? n d ? i d 1 2 \sum d = 1 n d 2 ? φ (d) + e (d) \sum i = 1 ? n d ? i

$\begin{align}& \sum_{i=1}^nf(i) \\= & \frac 1 2 \sum_{d=1}^n d\cdot \varphi(d)+e(d)\sum_{i=1}^{\lfloor\frac n d\rfloor} id \\= & \frac 1 2\sum_{d=1}^n d^2\cdot\varphi(d)+e(d)\sum_{i=1}^{\lfloor\frac nd\rfloor} i \\ \end{align}$
我们只需要杜教筛求出

id2?φ $id^2\cdot\varphi$ 的前缀和就可以了。

(id2?φ)×id2=id2?(φ×1)=id3 $(id^2\cdot\varphi)\times id^2=id^2\cdot(\varphi\times 1)=id^3$
因此

= \sum i = 1 n i 2 φ (i) \sum i = 1 n i 3 ? \sum i = 2 n i 2 \sum d = 1 ? n i ? d 2 φ (d)

$\begin{align} & \sum_{i=1}^n i^2\varphi(i)\\ = & \sum_{i=1}^n i^3-\sum_{i=2}^ni^2\sum_{d=1}^{\lfloor\frac n i\rfloor}d^2\varphi(d) \end{align}$
求出

f <script id="MathJax-Element-984" type="math/tex">f</script>的前缀和之后不难得到答案。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
const int p=1000000007,maxn=5000010,mod=1000007,inv2=500000004,inv6=166666668;
int phi[maxn],prm[maxn],sum[maxn],tot,m;
LL n;
/*void pause() {int x;x=1; }*/
int inc(int x,int y)
{x+=y;return x>=p?x-p:x;
}
int dec(int x,int y)
{x-=y;return x<0?x+p:x;
}
struct hashset
{int fir[mod+10],ne[mod*5],ans[mod*5],tot;LL val[mod*5];int find(LL n){for (int i=fir[n%mod];i;i=ne[i])if (val[i]==n) return ans[i];return -1;}void ins(LL n,int x){if (tot>mod*5-5) return;int y=n%mod;ne[++tot]=fir[y];fir[y]=tot;val[tot]=n;ans[tot]=x;}
}h;
int sum1(LL n)
{n%=p;return n*inc(n,1)%p*inv2%p;
}
int sum2(LL n)
{n%=p;return n*inc(n,1)%p*inc(inc(n,n),1)%p*inv6%p;
}
int sum3(LL n)
{int x=sum1(n);return (LL)x*x%p;
}
int calc(LL n)
{if (n<=m) return sum[n];int ret=h.find(n),ans,x,y;if (ret!=-1) return ret;ans=sum3(n);for (LL i=2,j;i<=n;i=j+1){//if (n==10000000000LL&&j==3333333333LL) pause();j=n/(n/i);x=dec(sum2(j),sum2(i-1));y=calc(n/i);ans=dec(ans,(LL)x*y%p);//if (x<0||y<0||ans<0) printf("no:%lld,%lld,%lld\n",n,i,j);}h.ins(n,ans);return ans;
}
int main()
{int ans=0,x,y;scanf("%lld",&n);m=pow(n,2.0/3);//m=100;phi[1]=sum[1]=1;for (int i=2;i<=m;i++){if (!phi[i]){prm[++tot]=i;phi[i]=i-1;}for (int j=1;j<=tot&&(LL)i*prm[j]<=m;j++)if (i%prm[j]) phi[i*prm[j]]=phi[i]*(prm[j]-1);else{phi[i*prm[j]]=phi[i]*prm[j];break;}sum[i]=inc(sum[i-1],(LL)phi[i]*i%p*i%p);}for (LL i=1,j;i<=n;i=j+1){j=n/(n/i);x=inc(dec(calc(j),calc(i-1)),i==1);y=sum1(n/i);ans=inc(ans,(LL)x*y%p);}printf("%d\n",dec(ans,sum1(n)));
}