线性筛+区间枚举

题目大意：给出区间a,b求区间内素数的个数

但是a,b非常大

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int N = 1e6 + 5;

bool vis[N], visab[N];
int prime[N], cnt = 0;
void is_prime()
{
	
	memset(vis, 0, sizeof(vis));
	vis[1] = 1;
	for (int i = 2; i<N; i++)
	{
		if (!vis[i])
		{
			prime[cnt++] = i;
		}
		for (int j = 0; j < cnt; j++) {
			if (i*prime[j] > N)break;
			vis[i*prime[j]] = 1;
			if (i%prime[j] == 0)break;//如果i能被已有的素数整数则跳出，避免重复标记
		}

	}
}


int main()
{
	//freopen("Text.txt", "r", stdin);
	int t;
	cin >> t;
	is_prime();
	for (int kase = 1; kase <= t; kase++)
	{
		LL a, b;
		scanf("%lld%lld", &a, &b);
		int count = 0;
		if (b <= N - 1)
		{
			for (LL i = a; i <= b; i++)
			{
				if (!vis[i])
					count++;
			}
		}
		else
		{
			memset(visab, 0, sizeof(visab));
			for (int i = 0; i<cnt&&prime[i] <= b; i++)
			{
				LL k = a / prime[i];
				if (k*prime[i]<a)
					k++;//求第一个使k*prime[i]>=a的倍数k
				for (LL j = k*prime[i]; j <= b; j += prime[i])
				{
					visab[j - a] = 1;//区间左移a
				}
			}
			for (LL i = a; i <= b; i++)
			{
				if (!visab[i - a])
					count++;
			}
		}
		printf("Case %d: %d\n", kase, count);
	}
}