51nod1584 加权约数和 莫比乌斯反演+线性筛

Solution

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求

$$\sum_{i=1}^n\sum_{j=1}^n\max(i,j)\cdot \sigma(i\cdot j)$$

Solution

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写完了回头看发现还是比较套路的，第一次写八级题肥肠激动
首先考虑去掉max

$$ans=2\sum_{i=1}^n\sum_{j=1}^i i\cdot \sigma(i\cdot j)-\sum_{i=1}^ni\cdot\sigma(i^2)$$
我们知道 $\sigma(n\cdot m)=\sum_{i|n}\sum_{j|m}\left[gcd(i,j)=1\right]\frac{n\cdot j}{m}$
然后此处略去一波套路得到 $$=\sum_{d=1}^n\mu(d)\cdot d^2\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sigma(i)\cdot i\sum_{j=1}^{i}\sigma(j)$$
这里就能用分块做了，然鹅根号的复杂度不足以跑过这题，考虑接着推
令 $T=id$，可以得到 $$=\sum_{T=1}^n\sum_{d|T}\mu(d)\cdot t\cdot T\cdot\sigma(\frac{T}{d})\sum_{i=1}^{\frac{T}{d}}\sigma(i)$$

大概就是介样，然后就可以枚举d然后nln预处理O(1)出解。注意不要忘了减去的那一部分

Code

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#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (register LL i=st;i<=ed;++i)

typedef long long LL;
const int MOD=1000000007;
const int N=1000005;

LL low[N+5],d[N+5],dd[N+5],mu[N+5];
LL ans[N+5],s[N+5],tmp[N+5];
LL prime[N/5];

bool not_prime[N+5];

int read() {
    int x=0,v=1; char ch=getchar();
    for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
    for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
    return x*v;
}

void add(LL &x,LL v) {
    x+=v; x-=(x>=MOD)?(MOD):(0);
}

void pre_work(int n) {
    dd[1]=d[1]=mu[1]=1;
    rep(i,2,n) {
        if (!not_prime[i]) {
            prime[++prime[0]]=i;
            mu[i]=-1; d[i]=i+1;
            low[i]=i; dd[i]=(i*i%MOD+i+1)%MOD;
        }
        for (int j=1;i*prime[j]<=n&&j<=prime[0];++j) {
            int x=i*prime[j];
            not_prime[x]=1;
            if (i%prime[j]==0) {
                mu[x]=0;
                low[x]=low[i]*prime[j];
                d[x]=(d[i]*prime[j]%MOD+d[i/low[i]])%MOD;
                dd[x]=(dd[i]*prime[j]%MOD*prime[j]%MOD+dd[i/low[i]]*prime[j]%MOD+dd[i/low[i]])%MOD;
                break;
            }
            low[x]=prime[j];
            d[x]=d[i]*d[prime[j]]%MOD;
            dd[x]=dd[i]*dd[prime[j]]%MOD;
            mu[x]=-mu[i];
        }
    }
    rep(i,1,n) s[i]=(s[i-1]+d[i])%MOD;
    for (LL i=1;i<=n;i++) {
        for (LL j=i;j<=n;j+=i) {
            LL tmp=mu[i]*i%MOD*j%MOD*d[j/i]%MOD*s[j/i]%MOD;
            add(ans[j],tmp);
        }
    }
    rep(i,1,n) add(ans[i],ans[i-1]);
    rep(i,1,n) add(tmp[i],(tmp[i-1]+i*dd[i]%MOD)%MOD);
}

int main(void) {
    freopen("data.in","r",stdin);
    freopen("myp.out","w",stdout);
    pre_work(1000000);
    for (int T=read(),n,cnt=0;T--;) {
        n=read();
        LL prt=(ans[n]*2%MOD-tmp[n]+MOD)%MOD;
        prt=(prt+MOD)%MOD;
        printf("Case #%d: %lld\n", ++cnt, prt);
    }
    return 0;
}