• BZOJ 4804: 欧拉心算


    数论题不多BB,直接开始推导吧:

    (sum_{i=1}^n sum_{j=1}^n phi(gcd(i,j)))

    (=sum_{i=1}^n sum_{j=1}^n sum_{d=1}^n [gcd(i,j)=d]phi(d))

    (=sum_{d=1}^n phi(d)cdot(sum_{i=1}^{lfloorfrac{n}{d} floor} sum_{j=1}^{lfloorfrac{n}{d} floor} [gcd(i,j)=1]))

    然后(sum_{i=1}^{lfloorfrac{n}{d} floor} sum_{j=1}^{lfloorfrac{n}{d} floor} [gcd(i,j)=1]=2cdotsum_{i=1}^{lfloorfrac{n}{d} floor} phi(i) -1),这个也是一个比较常用的结论了吧。具体可以看Luogu P2568 GCD的推导,之后为了方便,我们记(sum_x=sum_{i=1}^x phi(i))

    则上式(=sum_{d=1}^n(phi(d)cdot (2cdot sum_{lfloorfrac{n}{d} floor}-1)))

    (=2cdot sum_{d=1}^n(phi(d)cdot sum_{lfloorfrac{n}{d} floor})-sum_n)

    接下来就好做了,我们先把欧拉函数求前缀和,然后对于询问除法分块处理不同的(sum_{lfloorfrac{n}{d} floor})即可。

    复杂度(O(n+Tsqrt n)),轻松跑过

    CODE

    #include<cstdio>
    #define RI register int
    using namespace std;
    const int P=1e7;
    int prime[P+5],phi[P+5],cnt,n,t; long long ans,sum[P+5]; bool vis[P+5];
    inline void resolve(void)
    {
        vis[1]=sum[1]=phi[1]=1; for (RI i=2;i<=P;++i)
        {
            if (!vis[i]) prime[++cnt]=i,phi[i]=i-1;
            for (RI j=1;j<=cnt&&i*prime[j]<=P;++j)
            {
                vis[i*prime[j]]=1; if (i%prime[j]) phi[i*prime[j]]=phi[i]*(prime[j]-1);
                else { phi[i*prime[j]]=phi[i]*prime[j]; break; }
            }
            sum[i]=sum[i-1]+phi[i];
        }
    }
    int main()
    {
        for (scanf("%d",&t),resolve();t;--t)
        {
            scanf("%d",&n); ans=0; RI i,l,r;
            for (l=1;l<=n;l=r+1) r=n/(n/l),ans+=sum[n/l]*(sum[r]-sum[l-1]);
            printf("%lld
    ",(ans<<1)-sum[n]);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/cjjsb/p/9720241.html
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