HDU 1695 容斥

又是求gcd=k的题，稍微有点不同的是，(i,j)有偏序关系，直接分块好像会出现问题，还好数据规模很小，直接暴力求就行了。

/** @Date    : 2017-09-15 18:21:35
  * @FileName: HDU 1695 容斥 或 莫比乌斯反演.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

LL pri[N];
LL mu[N];
LL sum[N];
int c = 0;
bool vis[N];

void prime()
{
	MMF(vis);
	MMF(sum);
	mu[1] = 1;
	for(int i = 2; i < N; i++)
	{
		if(!vis[i])
			pri[c++] = i, mu[i] = -1;
		for(int j = 0; j < c && i * pri[j] < N; j++)
		{
			vis[i * pri[j]] = 1;
			if(i % pri[j] == 0)
			{
				mu[i * pri[j]] = 0;
				break;
			}
			else mu[i * pri[j]] = -mu[i];
		}
	}
	sum[0] = 0;
	for(int i = 1; i < N; i++)
		sum[i] += sum[i - 1] + mu[i];
}

LL get_sum(LL n, LL m)
{
	if(n > m) swap(n, m);
	int mi = min(n, m);
	LL ans = 0;
	for(int i = 1, last; i <= mi; i++, last = last + 1)
	{
		last = min(n/(n/i), m/(m/i));//由于有重复情况 不能直接分块？
		ans += (LL)(n / i) * (m / i) * (sum[i] - sum[i - 1]);
	}
	return ans;
}

int main()
{
	int T;
	prime();
	cin >> T;
	int cnt = 0;
	while(T--)
	{
		LL a, b, c, d, k;
		scanf("%lld%lld%lld%lld%lld", &a, &b, &c, &d, &k);
		if(k == 0)
		{
			printf("Case %d: 0\n", ++cnt);
			continue;
		}
		a = (a - 1) / k;
		b = b / k;
		c = (c - 1) / k;
		d = d / k;
		LL ans = get_sum(a, c) + get_sum(b, d) - get_sum(a, d) - get_sum(b, c) - get_sum(min(b,d), min(b,d)) / 2;
		printf("Case %d: %lld\n", ++cnt, ans);
		/*LL ans = 0;
		LL t = 0;
		for(int i = 1; i <= b; i++)
			ans += (b / i) * (d / i) * mu[i];
		for(int i = 1; i <= d; i++)
			t += (min(b,d)/ i) * (min(b, d) / i) * mu[i];
		printf("Case %d: %lld\n", ++cnt,  ans - t / 2);*/
	}
    return 0;
}