51Nod.1237.最大公约数之和 V3(莫比乌斯反演 杜教筛 欧拉函数)

题目链接

\(Description\)

　　\(n\leq 10^{10}\)，求
\[\sum_{i=1}^n\sum_{j=1}^ngcd(i,j)\ mod\ (1e9+7)\]

\(Solution\)

　　首先
\[\sum_{i=1}^n\sum_{j=1}^ngcd(i,j)=\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^n[gcd(i,j)=d]\]
　　注意不是\(\sum_{i=1}^n\sum_{j=1}^ngcd(i,j)=\sum_{d=1}^n\sum_{i=1}^n\sum_{j=1}^n[gcd(i,j)=d]\)！
\[ \begin{aligned} \sum_{i=1}^n\sum_{j=1}^ngcd(i,j)&=\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^n[gcd(i,j)=d]\\ &=\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}[gcd(i,j)=1]\\ &=\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\mu(i)\lfloor\frac{\lfloor\frac{n}{d}\rfloor}{i}\rfloor\lfloor\frac{\lfloor\frac{n}{d}\rfloor}{i}\rfloor\\ &=\sum_{d=1}^nd\sum_{d|t,t\leq n}\mu(\frac{t}{d})(\lfloor\frac{n}{t}\rfloor)^2\\ &=\sum_{t=1}^n(\lfloor\frac{n}{t}\rfloor)^2\sum_{d|t}d\mu(\frac{t}{d})\\ &=\sum_{t=1}^n(\lfloor\frac{n}{t}\rfloor)^2\varphi(t) \end{aligned} \]
　　然后就可以杜教筛了。
　　最后一步用到\[\sum_{d|n}\frac{n}{d}\mu(d)=\varphi(n)\]

证明：
　　首先有 \(\sum_{d|n}\varphi(d)=n\)（不证了）.
　　设 \(f(n)=n\)，则\(f(n)=\sum_{d|n}\varphi(d)\).
　　那么 \(\varphi(n)=\sum_{d|n}f(\frac{n}{d})\mu(d)\)
　　即 \(\sum_{d|n}\frac{n}{d}\mu(d)=\varphi(n)\).

//前缀和不要忘取模。。
#include<cstdio>
#include<cstring>
typedef long long LL;
const int N=4500000,mod=1e9+7;

int P[N>>2],cnt,phi[N+3];
LL n,sum[N+3],sum2[100000];
bool Not_P[N+3];

void Init()
{
    phi[1]=1;
    for(int i=2;i<=N;++i)
    {
        if(!Not_P[i]) P[++cnt]=i,phi[i]=i-1;
        for(int j=1;j<=cnt&&i*P[j]<=N;++j)
        {
            Not_P[i*P[j]]=1;
            if(i%P[j]) phi[i*P[j]]=phi[i]*(P[j]-1);
            else {phi[i*P[j]]=phi[i]*P[j]; break;}
        }
    }
    for(int i=1;i<=N;++i) sum[i]=sum[i-1]+phi[i], sum[i]>=mod?sum[i]-=mod:0;
}
LL FP(LL x,LL k)
{
    LL t=1;
    for(;k;k>>=1,x=x*x%mod)
        if(k&1) t=t*x%mod;
    return t;
}
const LL inv2=FP(2,mod-2);
LL Calc(LL x)
{
    if(x<=N) return sum[x];
    else if(~sum2[n/x]) return sum2[n/x];
    LL t=x%mod, res=t*(t+1)%mod*inv2%mod;
    for(LL i=2,las;i<=x;i=las+1)
        las=x/(x/i),(res-=(las-i+1)*Calc(x/i)%mod)%=mod;
    return sum2[n/x]=(res+mod)%mod;
}

int main()
{
    Init();//scanf("%I64d",&n);
    scanf("%lld",&n);
    memset(sum2,0xff,sizeof sum2);
    LL res=0;
    for(LL i=1,las,t;i<=n;i=las+1)
        las=n/(n/i), t=n/i%mod, (res+=t*t%mod*(Calc(las)-Calc(i-1)+mod)%mod)%=mod;
    printf("%lld",res);

    return 0;
}//9800581876

转载于:https://www.cnblogs.com/SovietPower/p/8412639.html