51nod 1594 Gcd and Phi 莫比乌斯反演

Description

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求

∑i=1n∑j=1nφ(gcd(φ(i),φ(j))) ∑ i = 1 n ∑ j = 1 n φ ( gcd ( φ ( i ) , φ ( j ) ) )

n<=2*10^6

Solution

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感觉还是数学题写得舒心(￣▽￣)”
令 s(d)=∑ni=1[φ(i)=d] s ( d ) = ∑ i = 1 n [ φ ( i ) = d ]
套路地枚举gcd得到

∑d=1nφ(d)∑i=1⌊nd⌋s(di)∑j=1⌊nd⌋s(dj)[gcd(i,j)=1] ∑ d = 1 n φ ( d ) ∑ i = 1 ⌊ n d ⌋ s ( d i ) ∑ j = 1 ⌊ n d ⌋ s ( d j ) [ g c d ( i , j ) = 1 ]

然后直接拉出来有

∑d=1nφ(d)∑x=1⌊nd⌋μ(x)⎛⎝∑i=1⌊ndx⌋s(dxi)⎞⎠2 ∑ d = 1 n φ ( d ) ∑ x = 1 ⌊ n d ⌋ μ ( x ) ( ∑ i = 1 ⌊ n d x ⌋ s ( d x i ) ) 2

令 f(x)=∑⌊nx⌋i=1s(xi) f ( x ) = ∑ i = 1 ⌊ n x ⌋ s ( x i )
这个f是可以nlogn求的，这整个柿子也是可以nlogn求的，直接在线做就可以了

注意到题目并没有给模数，因此这题是不用膜的（废话
Tnlogn的复杂度不太像能过，要卡卡才能过

Code

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#include 
#include 
#define rep(i,st,ed) for (register int i=st;i<=ed;++i)

typedef long long LL;
const int N=2000005;

LL f[N+5],phi[N+5];

int mu[N+5],prime[N+5],s[N+5];

bool not_prime[N+5];

void pre_work(int n) {
    phi[1]=mu[1]=1;
    rep(i,2,n) {
        if (!not_prime[i]) {
            prime[++prime[0]]=i;
            phi[i]=i-1; mu[i]=-1;
        }
        for (int j=1;i*prime[j]<=n&&j<=prime[0];++j) {
            int x=i*prime[j];
            not_prime[x]=true;
            if (i%prime[j]==0) {
                mu[x]=0; phi[x]=phi[i]*prime[j];
                break;
            }
            mu[x]=-mu[i]; phi[x]=phi[i]*(prime[j]-1);
        }
    }
}

void solve(int n) {
    rep(i,1,n) {
        s[i]=0;
        s[phi[i]]++;
    }
    rep(i,1,n) {
        f[i]=0;
        for (int j=i;j<=n;j+=i) {
            f[i]+=s[j];
        }
        f[i]*=f[i];
    }
    LL ans=0;
    rep(i,1,n) {
        LL tmp=0;
        for (int j=1;i*j<=n;++j) {
            tmp+=f[i*j]*mu[j];
        }
        ans=ans+tmp*phi[i];
    }
    printf("%lld\n", ans);
}

int main(void) {
    pre_work(N);
    int T; for (scanf("%d",&T);T--;) {
        int n; scanf("%d",&n);
        solve(n);
    }
    return 0;
}