51nod1238 最小公倍数之和 V3

题解

　　感觉柿子画起来有点费劲，我太蒻了。

ans=∑i=1n∑j=1nij(i,j)=∑d=1n1d∑d|in∑d|jnij[(i,j)=1]=∑d=1nd∑i=1⌊nd⌋∑j=1⌊nd⌋ij[(i,j)=1]=∑d=1nd×(2∑i=1⌊nd⌋i(∑j=1ij[(i,j)=1]))−1)=∑d=1nd(2∑i=1⌊nd⌋iiφ(i)+[i=1]2−1)=∑d=1nd∑i=1⌊nd⌋i2φ(i)

　　令f(i)=i2φ(i)，s(n)=∑ni=1f(i)

ans=∑d=1nd×s(⌊nd⌋)

　　id2(n)=n2

∑d|nd2φ(i)id2(nd)=∑d|nd2φ(i)n2d2=n3

　　要画出杜教筛，要试着对卷积求个前缀和

∑i=1n∑d|if(d)id2(id)=∑id=1n∑d=1⎢⎣⎢⎢nid⎥⎦⎥⎥f(d)id2(id)=∑t=1n∑d=1⌊nt⌋f(d)id2(t)=∑t=1nid2(t)s(⌊nt⌋)

　　因为强迫症，我把t都换成i。（和之前的i没有任何关系）

∑i=1nid2(i)s(⌊ni⌋)=∑i=1ni3s(n)=∑i=1ni3−∑i=2nid2(i)s((⌊ni⌋)

　　好了

代码

//杜教筛 #include <cstdio>#include <algorithm>#define maxn 4700000#define mod 1000000007llusing namespace std;typedef long long ll;ll f[maxn+10], phi[maxn+10], _2, _6, N;int prime[maxn+10];bool mark[maxn+10];void shai(){    ll i, j;    phi[1]=1;    for(i=2;i<maxn;i++)    {        if(!mark[i])prime[++*prime]=i, phi[i]=i-1;        for(j=1;i*prime[j]<maxn;j++)        {            mark[i*prime[j]]=1;            if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}            phi[i*prime[j]]=phi[i]*phi[prime[j]];        }    }    for(i=1;i<maxn;i++)phi[i]=(phi[i]*i%mod*i%mod+phi[i-1])%mod;}ll sqr(ll x){if(x>mod)x%=mod;return x*x%mod;}ll s1(ll x){if(x>mod)x%=mod;return x*(1+x)%mod*_2%mod;}ll s2(ll x){if(x>mod)x%=mod;return x*(1+x)%mod*(2*x+1)%mod*_6%mod;}ll s3(ll x){if(x>mod)x%mod;return sqr(s1(x));}ll getf(ll x){return x<maxn?phi[x]:f[N/x];}void djs(ll n){    if(n<maxn or getf(n))return;    ll i, last, t=N/n;    f[t]=s3(n);    for(i=2;i<=n;i=last+1)    {        last=n/(n/i);        djs(n/i);        f[t]=(f[t]-(s2(last)-s2(i-1))*getf(n/i)%mod)%mod;    }}int main(){    ll i, last, ans=0;    _2=500000004;    _6=166666668;    scanf("%lld",&N);    shai();    djs(N);    for(i=1;i<=N;i=last+1)    {        last=N/(N/i);        ans=(ans+getf(N/i)*(s1(last)-s1(i-1)))%mod;    }    printf("%lld",(ans+mod)%mod);    return 0;}