求第 i 个素数 Meissel Lehmer Algorithm + 二分 【模板】

1473: L先生与质数V3

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 1348  Solved: 147
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Description

在解决了上一个质数问题之后，L先生依然不甘心，他还想计算下更多范围内的质数，你能帮助他吗？

Input

有多组测试例。（测试例数量<70)
每个测试例一行，输入一个数字n（0<n<=3000000），输入0表示结束。

Output

输出测试例编号和第N个质数。
Case X: Y

Sample Input

1
2
3
4
10
100
0

Sample Output

Case 1: 2
Case 2: 3
Case 3: 5
Case 4: 7
Case 5: 29
Case 6: 541

每次二分判断即可；
（参考别人的代码）

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
typedef long long LL;
const int N = 5e6 + 2;//通过知道前面的n^1/3的=质数可以推断后面n^2/3的质数所以可以适当减小
bool np[N];
int prime[N], pi[N];
int getprime()
{
    int cnt = 0;
    np[0] = np[1] = true;
    pi[0] = pi[1] = 0;
    for(int i = 2; i < N; ++i)
    {
        if(!np[i]) prime[++cnt] = i;
        pi[i] = cnt;
        for(int j = 1; j <= cnt && i * prime[j] < N; ++j)
        {
            np[i * prime[j]] = true;
            if(i % prime[j] == 0)   break;
        }
    }
    return cnt;
}
const int M = 7;//为了减小内存可以不过是质数
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;//为了减小内存可以不过要按质数减小如去掉17
int phi[PM + 1][M + 1], sz[M + 1];
void init()
{
    getprime();
    sz[0] = 1;
    for(int i = 0; i <= PM; ++i)  phi[i][0] = i;
    for(int i = 1; i <= M; ++i)
    {
        sz[i] = prime[i] * sz[i - 1];
        for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];
    }
}
int sqrt2(LL x)
{
    LL r = (LL)sqrt(x - 0.1);
    while(r * r <= x)   ++r;
    return int(r - 1);
}
int sqrt3(LL x)
{
    LL r = (LL)cbrt(x - 0.1);
    while(r * r * r <= x)   ++r;
    return int(r - 1);
}
LL getphi(LL x, int s)
{
    if(s == 0)  return x;
    if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];
    if(x <= prime[s]*prime[s])   return pi[x] - s + 1;
    if(x <= prime[s]*prime[s]*prime[s] && x < N)
    {
        int s2x = pi[sqrt2(x)];
        LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;
        for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];
        return ans;
    }
    return getphi(x, s - 1) - getphi(x / prime[s], s - 1);
}
LL getpi(LL x)
{
    if(x < N)   return pi[x];
    LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;
    for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;
    return ans;
}
LL lehmer_pi(LL x)
{
    if(x < N)   return pi[x];
    int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
    int b = (int)lehmer_pi(sqrt2(x));
    int c = (int)lehmer_pi(sqrt3(x));
    LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2;
    for (int i = a + 1; i <= b; i++)
    {
        LL w = x / prime[i];
        sum -= lehmer_pi(w);
        if (i > c) continue;
        LL lim = lehmer_pi(sqrt2(w));
        for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);
    }
    return sum;
}
int main()//因为统计了质数个数可以用在线判断用二分
{
    LL n,l,r,mid;
    int sum=1;
    init();
    while(scanf("%lld",&n)!=EOF)
    {
       if(n==0)
        break;
       l=1,r=50000000;
       while(l<r) {
        mid=(l+r)/2;

        LL t=lehmer_pi(mid);
        if(t>=n)
            r=mid;
        else
            l=mid+1;
       }
       printf("Case %d: %lld\n",sum++,l);
    }
    return 0;
}