2019南京网络赛E：K Sum

Description:

定义函数

\[f _n (k) = \sum _{l _1 = 1} ^n \sum _{l _2 = 1} ^n \cdots \sum _{l _k = 1} ^n \gcd(l _1, l _2, \cdots, l _k) ^2 \]

现给定 \(n, k\)，需要求出 \(\sum _{i = 2} ^k f _n (i)\)，答案对 \(10 ^9 + 7\) 取模。

\(T\) 组数据。

\[1 \le T \le 10, 1 \le n \le 10 ^9, 2 \le k \le 10 ^{10 ^5} \]

Solution:

\[\begin{aligned} f_n(k) &= \sum_{d=1}^nd^2\sum_{x=1}^\frac{n}{d}\mu(x)\lfloor\frac{n}{dx}\rfloor ^k\\ &=\sum_{T=1}^n\lfloor\frac{n}{T}\rfloor^k\sum_{d|T}\mu(d)\left(\frac{T}{d}\right)^2 \end{aligned} \]

答案为:

\[\begin{aligned} &\sum_{i=2}^k\sum_{T=1}^n\lfloor\frac{n}{T}\rfloor^i\sum_{d|T}\mu(d)\left(\frac{T}{d}\right)^2\\ =&\sum_{T=1}^n\left(\sum_{i=2}^k\lfloor\frac{n}{T}\rfloor^i\right)\left(\sum_{d|T}\mu(d)\left(\frac{T}{d}\right)^2\right) \end{aligned} \]

前面整除分块+等比数列求和，后面杜教筛。

Code:

#include <iostream>
#include <vector>
#include <map>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <fstream>

typedef long long LL;
typedef unsigned long long uLL;

#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define MP(x, y) std::make_pair(x, y)
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define GO cerr << "GO" << endl;

using namespace std;

inline void proc_status()
{
	ifstream t("/proc/self/status");
	cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}

template<class T> inline T read() 
{
	register T x(0);
	register char c;
	register int f(1);
	while (!isdigit(c = getchar())) if (c == '-') f = -1;
	while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));
	return x * f;
}

template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }

const int maxN = 5e6;
const int mod = 1e9 + 7;

LL qpow(LL a, LL b)
{
	LL ans = 1;
	while (b)
	{
		if (b & 1) ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return ans;
}

LL k, kk;
LL calc(int x)
{
	if (x == 1) return (kk - 1 + mod) % mod;
	else return (LL)x * (qpow(x, k) - x + mod) % mod * qpow(x - 1, mod - 2) % mod;
}

void Read()
{
	k = kk = 0;
	register char c;
	while (!isdigit(c = getchar()));
	while (k = (k * 10 + (c xor 48)) % (mod - 1), kk = (kk * 10 + (c xor 48)) % mod, isdigit(c = getchar()));
}

vector<int> prime;
bool vis[maxN + 2];
int f[maxN + 2], inv6;
map<int, int> F;

void Init()
{
	f[1] = 1;
	for (register int i = 2; i <= maxN; ++i)
	{
		if (!vis[i])
		{
			f[i] = ((LL)i * i - 1) % mod;
			prime.push_back(i);
		}
		for (register int j = 0; j < SZ(prime) && prime[j] * i <= maxN; ++j)
		{
			int t = prime[j] * i;
			vis[t] = 1;
			if (i % prime[j] == 0)
			{
				f[t] = (LL)f[i] * prime[j] % mod * prime[j] % mod;
				break;
			}
			else
				f[t] = (LL)f[i] * f[prime[j]] % mod;
		}
	}
	for (register int i = 2; i <= maxN; ++i) 
		f[i] += f[i - 1], f[i] %= mod;
}

int Sum(int n)
{
	if (n <= maxN) return f[n];
	if (F.count(n)) return F[n];
	LL ans = (LL)n * (n + 1) % mod * (2ll * n + 1) % mod * inv6 % mod;
	for (int l = 2, r; l <= n; l = r + 1)
	{
		r = n / (n / l);
		ans += mod - (LL)Sum(n / l) * (r - l + 1) % mod;
		ans %= mod;
	}
	return F[n] = ans % mod;
}

int main() 
{
#ifndef ONLINE_JUDGE
	freopen("xhc.in", "r", stdin);
	freopen("xhc.out", "w", stdout);
#endif
	int T = read<int>();
	inv6 = qpow(6, mod - 2);
	Init();
	while (T--)
	{
		LL ans(0);
		int n;
		n = read<int>(), Read();
		for (int l = 1, r; l <= n; l = r + 1)
		{
			r = n / (n / l);
			ans += calc(n / r) * (Sum(r) - Sum(l - 1) + mod) % mod;
			ans %= mod;
		}
		printf("%d\n", (int) ans);
	}
	return 0;
}