1489: L先生与质数V4（二分+大区间求素数模板）

 题意

L先生想求出第n个质数（素数）是多少，你能帮助他吗？

数据

T < 70, 0 < n <= 3e6, 输入0表示结束。

输入

1

2

3

4

10

100

0

输出

Case 1: 2

Case 2: 3

Case 3: 5

Case 4: 7

Case 5: 29

Case 6: 541

HDU5901是个求1-1e11内素数个数的模板题，这题可以利用二分+那个模板求解。（用那的第二个板子应该会更快些，但不知道为什么迷之RE，所以只能将就套第一个）

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

#define MAXN 100
#define MAXM 10001
#define MAXP 40000
#define MAX 400000
#define clr(ar) memset(ar, 0, sizeof(ar))
#define read() freopen("lol.txt", "r", stdin)
#define dbg(x) cout << #x << " = " << x << endl
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))

using namespace std;
typedef long long ll;

namespace pcf{
    long long dp[MAXN][MAXM];
    unsigned int ar[(MAX >> 6) + 5] = {0};
    int len = 0, primes[MAXP], counter[MAX];

    void Sieve(){
        setbit(ar, 0), setbit(ar, 1);
        for (int i = 3; (i * i) < MAX; i++, i++){
            if (!chkbit(ar, i)){
                int k = i << 1;
                for (int j = (i * i); j < MAX; j += k) setbit(ar, j);
            }
        }

        for (int i = 1; i < MAX; i++){
            counter[i] = counter[i - 1];
            if (isprime(i)) primes[len++] = i, counter[i]++;
        }
    }

    void init(){
        Sieve();
        for (int n = 0; n < MAXN; n++){
            for (int m = 0; m < MAXM; m++){
                if (!n) dp[n][m] = m;
                else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];
            }
        }
    }

    long long phi(long long m, int n){
        if (n == 0) return m;
        if (primes[n - 1] >= m) return 1;
        if (m < MAXM && n < MAXN) return dp[n][m];
        return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);
    }

    long long Lehmer(long long m){
        if (m < MAX) return counter[m];

        long long w, res = 0;
        int i, a, s, c, x, y;
        s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);
        a = counter[y], res = phi(m, a) + a - 1;
        for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;
        return res;
    }
}

long long solve(long long n){
    int i, j, k, l;
    long long x, y, res = 0;

    for (i = 0; i < pcf::len; i++){
        x = pcf::primes[i], y = n / x;
        if ((x * x) > n) break;
        res += (pcf::Lehmer(y) - pcf::Lehmer(x));
    }

    for (i = 0; i < pcf::len; i++){
        x = pcf::primes[i];
        if ((x * x * x) > n) break;
        res++;
    }

    return res;
}

int main()
{
    pcf::init();
    ll n, ca = 1;
    while(cin >> n && n)
    {
        ll l = 2, r = 1e8, ans;
        while(l <= r)
        {
            ll mid = (l+r)/2;
            if(pcf::Lehmer(mid) >= n)
                r = mid-1, ans = mid;
            else l = mid+1;
        }
        printf("Case %lld: %lld\n", ca++, ans);
    }
    return 0;
}