HDU 4135 容斥

问a,b区间内与n互质个数，a,b<=1e15,n<=1e9

n才1e9考虑分解对因子的组合进行容斥，因为19个最小的不同素数乘积即已大于LL了，枚举状态复杂度不会很高。然后差分就好了。

/** @Date    : 2017-09-28 16:52:30
  * @FileName: HDU 4135 容斥.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];
bool vis[N];
int c;
void prime()
{
	MMF(vis);
	for(int i = 2; i < N; i++)
	{
		if(!vis[i]) pri[c++] = i;
		for(int j = 2; j < c && i * pri[j] < N; j++)
		{
			vis[i * pri[j]] = 1;
			if(i % pri[j] == 0) break;
		}
	}
	//19个最小的不同素数乘积即已大于2^63 - 1
	/*LL t = 1;
	for(int i = 0; i < 100; i++)
		t *= pri[i], cout << t <<"~" <<i<<endl;*/
}

int main()
{
	prime();
	int T;
	cin >> T;
	int icas = 0;
	while(T--)
	{
		LL n, a, b;
		scanf("%lld%lld%lld", &a, &b, &n);

		int cnt = 0;
		vector<LL>q;
		for(int i = 0; i < c && pri[i] <= n / pri[i]; i++)
		{
			if(n % pri[i] == 0)
			{
				while(n % pri[i] == 0)
					n /= pri[i];
				q.PB(pri[i]);
				cnt++;
			}
		}
		if(n > 1)
			q.PB(n), cnt++;
		LL ansa = 0;
		LL ansb = 0;
		for(int i = 1; i < (1 << cnt); i++)
		{
			LL t = 1;
			int f = -1;
			for(int j = 0; j < cnt; j++)
				if((i & (1 << j)))
					t *= q[j], f *= -1;
			//cout << t << endl;
			ansb += f * (b / t);
			ansa += f * ((a - 1) / t);
		}
		//cout << ansa <<"~" <<ansb<<endl;
		LL ans = b - ansb - (a - 1 - ansa);
		printf("Case #%d: %lld\n", ++icas, ans);
	}
    return 0;
}