【数论】Prime Time UVA - 10200 大素数 Miller Robin 模板

题意：验证1~10000 的数 n^n+n+41 中素数的个数。每个询问给出a,b 　求区间[a,b]中质数出现的比例，保留两位

题解：质数会爆到1e8 所以用miller robin ，

另外一个优化是预处理

一个坑是四舍五入卡精度。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<ctime>
using namespace std;
typedef long long ll;
const int MAXN =  +  + ;
const int maxn = MAXN;
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
int phi[MAXN], prime[MAXN];
struct Miller_Rabin
{
    int prime[] = { ,,,, };
    ll qmul(ll x, ll y, ll mod) {
        ll ans = (x*y - (ll)((long double)x / mod * y + 1e-)*mod);
        ans = (ans%mod + mod) % mod;
        return ans;
    }
    ll qpow(ll x, ll n, ll mod) {
        ll ans = ;
        while (n) {
            if (n & ) ans = qmul(ans, x, mod);
            x = qmul(x, x, mod);
            n >>= ;
        }
        return ans;
    }
    bool isprime_std(ll p) {
        if (p < ) return ;
        if (p !=  && p %  == ) return ;
        ll s = p - ;
        while (!(s & )) s >>= ;
        for (int i = ; i < ; ++i) {
            if (p == prime[i]) return ;
            ll t = s, m = qpow(prime[i], s, p);
            while (t != p -  && m !=  && m != p - ) {
                m = qmul(m, m, p);
                t <<= ;
            }
            if (m != p -  && !(t & )) return ;
        }
        return ;
    }
    bool isprime(ll p) {
        if (p< || (p !=  && p %  == )) return false;
        for (int i = ; i < ; ++i)
        {
            if (p == prime[i]) return true;
            ll t = qpow(prime[i], p - , p);
            if (t != ) return false;
        }
        return true;
    }
}mr;
int tot;
void get_phi()
{
    phi[] = ;
    for (int i = ; i <= MAXN - ; i++) {
        if (!phi[i]) {
            phi[i] = i - ;
            prime[++tot] = i;
        }
        for (int j = ; j <= tot && 1LL * i*prime[j] <= MAXN - ; j++) {
            if (i%prime[j]) phi[i*prime[j]] = phi[i] * (prime[j] - );
            else {
                phi[i*prime[j]] = phi[i] * prime[j];
                break;
            }
        }
    }
}
int isntp[maxn];
void sieve(int n) {
    int m = (int)sqrt(n + 0.5);
    mmm(isntp, );
    rep(i, , m)if (!isntp[i])for (int j = i * i; j <= n; j += i)isntp[j] = ;

}
int ans[maxn];
int s[maxn];
int smain();
//#define ONLINE_JUDGE
int main() {

    //ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    FILE *myfile;
    myfile = freopen("C:\\Users\\acm-14\\Desktop\\test\\b.in", "r", stdin);
    if (myfile == NULL)
        fprintf(stdout, "error on input freopen\n");
    FILE *outfile;
    outfile = freopen("C:\\Users\\acm-14\\Desktop\\test\\out.txt", "w", stdout);
    if (outfile == NULL)
        fprintf(stdout, "error on output freopen\n");
    long _begin_time = clock();
#endif
    smain();
#ifndef ONLINE_JUDGE
    long _end_time = clock();
    printf("time = %ld ms.", _end_time - _begin_time);
#endif
    return ;
}
int smain()
{

    int t;
    int a, b;

    s[] = ;
    rep(i, , 1e4) {
        if (mr.isprime_std(i * i + i + ))s[i] = s[i - ] + ;
        else s[i] = s[i - ];
    }
    while (cin >> a >> b)
    {
        int cnt = ;
        /*rep(i, a, b) {
        if (i * i + i + 41 < 1e7) {
        if (isntp[i * i + i + 41] == 0)cnt++;
        else if(mr.isprime(i * i + i + 41))cnt++;
        }
        }*/
        cnt = s[b];
        if (a != )cnt -= s[a - ];
        double ans = (double)cnt / (double)(b - a + ) * ;
        ans = (double)((int)(ans + 0.50000001));

        printf("%.2lf\n", ans/);
    }
    //cin >> t;
    return ;
}
/*
0 39
0 40
39 40
*/