luogu3768 简单的数学题

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题意：(sum_{i=1}^nsum_{j=1}^nijgcd(i,j))

对输入的某素数取模，一组询问，(nle10^{10})，最大一组时限6s

推式子(orz某题解推了三行。。。)

(sum_{i=1}^nsum_{j=1}^nijgcd(i,j))

(=sum_{p=1}^npsum_{i=1}^nsum_{j=1}^nij[gcd(i,j)=p])

(=sum_{p=1}^np^3sum_{i=1}^{n/p}sum_{j=1}^{n/p}ij[gcd(i,j)=1])

(=sum_{p=1}^np^3sum_{i=1}^{n/p}sum_{j=1}^{n/p}ijsum_{d|i,d|j}mu(d))

(=sum_{p=1}^np^3sum_{d=1}^nmu(d)d^2sum_{i=1}^{n/pd}sum_{j=1}^{n/pd}ij)

(=sum_{q=1}^nq^2(sum_{d|q}mu(d)frac qd)(sum_{i=1}^{n/q}i)^2)

(=sum_{q=1}^nq^3(sum_{d|q}frac {mu(d)}d)(sum_{i=1}^{n/q}i)^2)

(=sum_{q=1}^nq^3(frac{varphi(q)}q)(sum_{i=1}^{n/q}i)^2)

(=sum_{q=1}^nq^2varphi(q)(sum_{i=1}^{n/q}i)^2)

整除分块+杜教筛(类似loj这是一道简单的数学题)

具体还是找个小样例跑一下就行了...

#include <cstdio>
using namespace std;

#define int long long

int n, p, prime[3000010], tot, phi[3000010], cjh = 3000000;
bool vis[3000010];

int qpow(int x, int y)
{
	int res = 1;
	while (y > 0)
	{
		if (y & 1) res = res * (long long)x % p;
		x = x * (long long)x % p;
		y >>= 1;
	}
	return res;
}

int inv2, inv6;

int s1(int x) { x %= p; return x * (long long)(x + 1) % p * inv2 % p; }
int s2(int x) { x %= p; return x * (long long)(x + 1) % p * (x * 2 + 1) % p * inv6 % p; }
int s3(int x) { return qpow(s1(x), 2); }

bool count[3000010];
int ans[3000010];

int chuans(int x)
{
	if (x <= cjh) return phi[x];
	if (count[n / x]) return ans[n / x];
	count[n / x] = true;
	int res = s3(x);
	for (int i = 2, j; i <= x; i = j + 1)
		j = (x / (x / i)), res = ((res - ((s2(j) - s2(i - 1)) * (long long)chuans(x / i) % p)) % p + p) % p;
	return ans[n / x] = res;
}

signed main()
{
	scanf("%lld%lld", &p, &n), phi[1] = 1;
	inv2 = qpow(2, p - 2), inv6 = qpow(6, p - 2);
	for (int i = 2; i <= cjh; i++)
	{
		if (vis[i] == false) prime[++tot] = i, phi[i] = i - 1;
		for (int j = 1; j <= tot && i * prime[j] <= cjh; j++)
		{
			vis[i * prime[j]] = true;
			if (i % prime[j] == 0) { phi[i * prime[j]] = phi[i] * prime[j]; break; }
			phi[i * prime[j]] = phi[i] * (prime[j] - 1);
		}
		phi[i] = phi[i] * (long long)i % p * i % p;
		phi[i] = (phi[i] + phi[i - 1]) % p;
	}
	int res = 0;
	for (int i = 1, j; i <= n; i = j + 1)
		j = (n / (n / i)), res = ((res + ((s3(n / i) - s3(n / (j + 1))) * (long long)chuans(j)) % p) % p + p) % p;
	printf("%lld
", res);
	return 0;
}

不define int long long莫名WA了两个点。。