原题题面

Given you n,x,k , find the value of the following formula:
∑ a 1 = 1 n ∑ a 2 = 1 n . . . ∑ a x = 1 n ( ∏ j = 1 x a j k ) f ( g c d ( a 1 , a 2 , . . . a x ) ) ∗ g c d ( a 1 , a 2 , . . . a x ) \sum_{a_1=1}^{n}\sum_{a_2=1}^{n}...\sum_{a_x=1}^{n}(\prod_{j=1}^{x}a_{j}^{k})f(gcd(a_1,a_2,...a_x))*gcd(a_1,a_2,...a_x) ∑a1=1n∑a2=1n...∑ax=1n(∏j=1xajk)f(gcd(a1,a2,...ax))∗gcd(a1,a2,...ax)
g c d ( a 1 , a 2 , … , a n ) gcd(a_1,a_2,…,a_n) gcd(a1,a2,…,an) is the greatest common divisor of a 1 , a 2 , . . . , a − n a_1,a_2,...,a-n a1,a2,...,a−n.
The function f ( x ) f(x) f(x) is defined as follows:
If there exists an ingeter k k k ( k > 1 ) (k>1) (k>1) , and k 2 k^2 k2 is a divisor of x x x,
then f ( x ) = 0 f(x)=0 f(x)=0, else f ( x ) = 1 f(x)=1 f(x)=1.

输入格式

The first line contains three integers t , k , x ( 1 ≤ t ≤ 1 0 4 , 1 ≤ k ≤ 1 0 9 , 1 ≤ x ≤ 1 0 9 ) t,k,x (1≤t≤10^4,1≤k≤10^9,1≤x≤10^9) t,k,x(1≤t≤104,1≤k≤109,1≤x≤109)
Then t t t test cases follow. Each test case contains an integer n n n ( 1 ≤ n ≤ 2 × 1 0 5 ) (1≤n≤2×10^5) (1≤n≤2×105)

输出格式

For each test case, print one integer — the value of the formula.
Because the answer may be very large, please output the answer modulo 1 0 9 + 7 10^9+7 109+7.

输入样例

3 1 3
56
5
20

输出样例

139615686
4017
11554723

题面分析

设 g c d ( a 1 , a 2 , . . . a x ) = d gcd(a_1,a_2,...a_x)=d gcd(a1,a2,...ax)=d，得
∑ a 1 = 1 n ∑ a 2 = 1 n . . . ∑ a x = 1 n ( ∏ j = 1 x a j k ) f ( d ) ∗ d [ g c d ( a 1 , a 2 , . . . a x ) = d ] \sum_{a_1=1}^{n}\sum_{a_2=1}^{n}...\sum_{a_x=1}^{n}(\prod_{j=1}^{x}a_{j}^{k})f(d)*d[gcd(a_1,a_2,...a_x)=d] ∑a1=1n∑a2=1n...∑ax=1n(∏j=1xajk)f(d)∗d[gcd(a1,a2,...ax)=d]
枚举 d d d，得到
∑ d = 1 n d ∗ f ( d ) ∗ ∑ a 1 = 1 n ∑ a 2 = 1 n . . . ∑ a x = 1 n ( ∏ j = 1 x a j k ) [ g c d ( a 1 , a 2 , . . . a x ) = d ] \sum_{d=1}^{n}d*f(d)*\sum_{a_1=1}^{n}\sum_{a_2=1}^{n}...\sum_{a_x=1}^{n}(\prod_{j=1}^{x}a_{j}^{k})[gcd(a_1,a_2,...a_x)=d] ∑d=1nd∗f(d)∗∑a1=1n∑a2=1n...∑ax=1n(∏j=1xajk)[gcd(a1,a2,...ax)=d]
∑ d = 1 n d k x + 1 ∗ f ( d ) ∗ ∑ a 1 = 1 ⌊ n d ⌋ ∑ a 2 = 1 ⌊ n d ⌋ . . . ∑ a x = 1 ⌊ n d ⌋ ( ∏ j = 1 x a j k ) [ g c d ( a 1 , a 2 , . . . a x ) = 1 ] \sum_{d=1}^{n}d^{kx+1}*f(d)*\sum_{a_1=1}^{\lfloor \frac{n}{d}\rfloor}\sum_{a_2=1}^{\lfloor \frac{n}{d}\rfloor}...\sum_{a_x=1}^{\lfloor \frac{n}{d}\rfloor}(\prod_{j=1}^{x}a_{j}^{k})[gcd(a_1,a_2,...a_x)=1] ∑d=1ndkx+1∗f(d)∗∑a1=1⌊dn⌋∑a2=1⌊dn⌋...∑ax=1⌊dn⌋(∏j=1xajk)[gcd(a1,a2,...ax)=1]
将 [ g c d ( a 1 , a 2 , . . . a x ) = 1 ] [gcd(a_1,a_2,...a_x)=1] [gcd(a1,a2,...ax)=1]化作 μ \mu μ，得到
∑ d = 1 n d k x + 1 ∗ f ( d ) ∗ ∑ a 1 = 1 ⌊ n d ⌋ ∑ a 2 = 1 ⌊ n d ⌋ . . . ∑ a x = 1 ⌊ n d ⌋ ( ∏ j = 1 x a j k ) ∑ t ∣ a 1 , t ∣ a 2 , . . . t ∣ a x μ ( t ) \sum_{d=1}^{n}d^{kx+1}*f(d)*\sum_{a_1=1}^{\lfloor \frac{n}{d}\rfloor}\sum_{a_2=1}^{\lfloor \frac{n}{d}\rfloor}...\sum_{a_x=1}^{\lfloor \frac{n}{d}\rfloor}(\prod_{j=1}^{x}a_{j}^{k})\sum_{t|a_1,t|a_2,...t|a_x}\mu(t) ∑d=1ndkx+1∗f(d)∗∑a1=1⌊dn⌋∑a2=1⌊dn⌋...∑ax=1⌊dn⌋(∏j=1xajk)∑t∣a1,t∣a2,...t∣axμ(t)
∑ d = 1 n d k x + 1 ∗ f ( d ) ∗ ( ∑ i = 1 ⌊ n d ⌋ i k ) x ∗ ∑ t ∣ a 1 , t ∣ a 2 , . . . t ∣ a x μ ( t ) \sum_{d=1}^{n}d^{kx+1}*f(d)*(\sum_{i=1}^{\lfloor \frac{n}{d}\rfloor}i^k)^x*\sum_{t|a_1,t|a_2,...t|a_x}\mu(t) ∑d=1ndkx+1∗f(d)∗(∑i=1⌊dn⌋ik)x∗∑t∣a1,t∣a2,...t∣axμ(t)
枚举 t t t得
∑ d = 1 n d k x + 1 ∗ t k x ∗ f ( d ) ∗ ∑ t = 1 ⌊ n d ⌋ μ ( t ) ( ∑ i = 1 ⌊ n d t ⌋ i k ) x \sum_{d=1}^{n}d^{kx+1}*t^{kx}*f(d)*\sum_{t=1}^{\lfloor \frac{n}{d}\rfloor}\mu(t)(\sum_{i=1}^{\lfloor \frac{n}{dt}\rfloor}i^k)^x ∑d=1ndkx+1∗tkx∗f(d)∗∑t=1⌊dn⌋μ(t)(∑i=1⌊dtn⌋ik)x
设 T = t d T=td T=td，得到
∑ T = 1 n T k x d f ( d ) ∗ ( ∑ i = 1 ⌊ n T ⌋ i k ) x ∗ ∑ t ∣ T μ ( t ) \sum_{T=1}^{n}T^{kx}df(d)*(\sum_{i=1}^{\lfloor \frac{n}{T}\rfloor}i^k)^x*\sum_{t|T}\mu(t) ∑T=1nTkxdf(d)∗(∑i=1⌊Tn⌋ik)x∗∑t∣Tμ(t)
为了方便计算，我们把右边的 t ∣ T t|T t∣T改成 d ∣ T d|T d∣T，得到
∑ T = 1 n T k x ( ∑ i = 1 ⌊ n T ⌋ i k ) x ∗ ∑ d ∣ T μ ( d ) d f ( T d ) \sum_{T=1}^{n}T^{kx}(\sum_{i=1}^{\lfloor \frac{n}{T}\rfloor}i^k)^x*\sum_{d|T}\mu(d)df(\frac{T}{d}) ∑T=1nTkx(∑i=1⌊Tn⌋ik)x∗∑d∣Tμ(d)df(dT)
设 G ( x ) = ∑ d ∣ T μ ( d ) d f ( T d ) G(x)=\sum_{d|T}\mu(d)df(\frac{T}{d}) G(x)=∑d∣Tμ(d)df(dT)
首先 μ \mu μ可以 O ( n ) O(n) O(n)预处理， f f f也可以 O ( n l o g n ) O(nlogn) O(nlogn)预处理
那么G可以在 O ( n ) O(n) O(n)下预处理
对于 ( ∑ i = 1 ⌊ n T ⌋ i k ) x (\sum_{i=1}^{\lfloor \frac{n}{T}\rfloor}i^k)^x (∑i=1⌊Tn⌋ik)x，也可以在 O ( n l o g n ) O(nlogn) O(nlogn)下处理完
对于每一次查询，利用分块和预处理前缀和，只需要 O ( n ) O(\sqrt{n}) O(n )。
故总复杂度为 O ( n l o g n + T n ) O(nlogn+T\sqrt{n}) O(nlogn+Tn )

AC代码(2199ms)

#include <bits/stdc++.h>
using namespace std;
const long long maxn=2e5;
const long long mod=1e9+7;
bool check[maxn+10];//访问标记
int prime[maxn+10];//质数
int mu[maxn+10];//mu函数
int f[maxn+10];//f函数
long long powk[maxn+10];
long long G[maxn+10];
long long sumG[maxn+10];
long long n, k, x;
inline long long quick_pow(long long a, long long b)//快速幂
{long long ans=1, base=a;while(b!=0){if (b&1)ans=(long long) ans*base%mod;base=(long long) base*base%mod;b>>=1;}return ans;
}
void init()
{f[1]=mu[1]=1;int tot=0;for(int i=2; i<=maxn; i++){f[i]=1;if (!check[i]){prime[tot++]=i;mu[i]=-1;}for(int j=0; j<tot; j++){if (i*prime[j]>maxn) break;check[i*prime[j]]=true;if (i%prime[j]==0){mu[i*prime[j]]=0;break;}else{mu[i*prime[j]]=-mu[i];}}}for(long long d=2; d<=maxn; d++)//处理f{for(long long j=d*d; j<=maxn; j+=d*d){f[j]=0;}}G[0]=0;for(long long d=1; d<=maxn; d++)//处理G{for(long long i=d; i<=maxn; i+=d){G[i]=(G[i]+d*f[d]%mod*mu[i/d]%mod+mod)%mod;}}powk[0]=0;sumG[0]=0;for(long long i=1; i<=maxn; i++)//处理累加{long long z=quick_pow(i, k);powk[i]=(powk[i-1]+z)%mod;sumG[i]=(sumG[i-1]+quick_pow(z, x)*G[i]%mod)%mod;}
}
long long answer()
{long long ans=0;for(long long l=1, r; l<=n; l=r+1){r=n/(n/l);ans=(ans+(sumG[r]-sumG[l-1]+mod)%mod*quick_pow(powk[n/l], x)%mod)%mod;}return ans;
}
void solve()
{int t;scanf("%d%lld%lld", &t, &k, &x);init();while(t--){scanf("%lld", &n);printf("%lld\n", answer());}
}
int main()
{//    ios_base::sync_with_stdio(false);
//    cin.tie(0);
//    cout.tie(0);
#ifdef ACM_LOCALfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);long long test_index_for_debug=1;char acm_local_for_debug;while(cin>>acm_local_for_debug){cin.putback(acm_local_for_debug);if (test_index_for_debug>100){throw runtime_error("Check the stdin!!!");}auto start_clock_for_debug=clock();solve();auto end_clock_for_debug=clock();cout<<"\nTest "<<test_index_for_debug<<" successful"<<endl;cerr<<"Test "<<test_index_for_debug++<<" Run Time: "<<double(end_clock_for_debug-start_clock_for_debug)/CLOCKS_PER_SEC<<"s"<<endl;cout<<"--------------------------------------------------"<<endl;}
#elsesolve();
#endifreturn 0;
}

DrGilbert 2020.8.7

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