给出多组a,b,c,d,ka,b,c,d,k，求

∑i=ab∑j=cd[gcd(i,j)==k] \sum_{i=a}^{b}\sum_{j=c}^{d}[\gcd(i,j)==k]

数据组数≤50000,a,b,c,d,k≤50000\le50000, a,b,c,d,k\le50000

Luogu P2522

Solution

一开始先容斥一下，就转化成「BZOJ1101」ZAP-Queries - 莫比乌斯反演 + 整除分块入门这道题了

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }

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

inline int read ()
{
	int sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

const int Maxn = 5e4 + 100;

int a, b, c, d, K;
int Prime[Maxn], Not_Prime[Maxn], prime_cnt, Mu[Maxn], Sum_Mu[Maxn];

inline LL Calc (int n, int m)
{
	n /= K, m /= K;
	int N = min(n, m);
	LL ans = 0;
	for (int l = 1, r; l <= N; l = r + 1)
	{
		r = min(n / (n / l), m / (m / l));
		ans += 1ll * (n / l) * (m / l) * (Sum_Mu[r] - Sum_Mu[l - 1]);
	}
	return ans;
}

inline void Solve ()
{
	printf("%lld\n", Calc(b, d) - Calc(b, c - 1) - Calc (a - 1, d) + Calc(a - 1, c - 1));
}

inline void Input ()
{
	a = read(), b = read(), c = read(), d = read(), K = read();
}

inline void Init ()
{
	Mu[1] = Not_Prime[1] = 1;
	for (int i = 2; i <= Maxn - 5; ++i)
	{
		if (!Not_Prime[i])
		{
			Prime[++prime_cnt] = i;
			Mu[i] = -1;
		}
		for (int j = 1; j <= prime_cnt && i * Prime[j] <= Maxn - 5; ++j)
		{
			Not_Prime[i * Prime[j]] = 1;
			if (!(i % Prime[j])) { Mu[i * Prime[j]] = 0; break; }
			else Mu[i * Prime[j]] = -Mu[i];
		}
	}
	for (int i = 1; i <= Maxn - 5; ++i) Sum_Mu[i] = Sum_Mu[i - 1] + Mu[i];
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("A.in", "r", stdin);
	freopen("A.out", "w", stdout);
#endif
	Init();
	int Testcase = read();
	while (Testcase--)
	{
		Input();
		Solve();
	}
	return 0;
}