用一個簡單的容斥就可以求出a<=x <=b, c<=y<=d範圍的答案了

注意1LL呀qwq

#include <bits/stdc++.h>
using namespace std;
#define N 50005
bool vis[N];
int miu[N],prime[N];
typedef long long LL;
LL sum[N],ans;
int a,b,c,d,k,T,cnt;
void Miu()
{
	miu[1] = 1;
	cnt = 0;
	for (int i=2;i<=N-5;i++)
	{
		if (!vis[i]) 
		{
			prime[++cnt] = i;
			miu[i] = -1;
		}
		for (int j = 1;j<=cnt && prime[j] * i <= N-5;j++)
		{
			vis[prime[j] * i] = 1;
			if (i % prime[j] == 0) break;
			else miu[i * prime[j]] = -miu[i];
		}
	}
	for (int i=1;i<=N-5;i++)
		sum[i] = sum[i-1] + (LL)miu[i];
}
LL get_ans(int n, int m)
{
	LL ans = 0;
	if (n > m) swap(n,m);
	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[r / k] - sum[(l-1)/k]);//注意long long
	}
	return ans;
}
int main()
{
	Miu();
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
		ans = get_ans(b,d) - get_ans(a-1,d) - get_ans(b,c-1) + get_ans(a-1,c-1);
		printf("%lld\n", ans);
	}
}