CodeForces 385C Bear and Prime Numbers 素数打表

第一眼看这道题目的时候觉得可能会很难也看不太懂，但是看了给出的Hint之后思路就十分清晰了

Consider the first sample. Overall, the first sample has 3 queries.



The first query l = 2, r = 11 comes. You need to count f(2) + f(3) + f(5) + f(7) + f(11) = 2 + 1 + 4 + 2 + 0 = 9.

The second query comes l = 3, r = 12. You need to count f(3) + f(5) + f(7) + f(11) = 1 + 4 + 2 + 0 = 7.

The third query comes l = 4, r = 4. As this interval has no prime numbers, then the sum equals 0.

xn的范围在(2 ≤ x_i ≤ 10⁷)，而 l, r 的范围在 (2 ≤ l_i ≤ r_i ≤ 2·10⁹) ，易得在计算的时候是不会考虑10⁷  以后了

首先写一个素数快速打表，同时也统计一下在 l, r 范围内每个数满足题目条件的总数 (虽然觉得这样的打表方法的确很慢)

然后注意了，因为查询的次数很多，多达5*10⁴次 ，所以在统计的时候可以算出累计和以节省时间

 

Source Code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <cstring>

#include <cmath>

#include <stack>

#include <string>

#include <map>

#include <set>

#include <list>

#include <queue>

#include <vector>

#include <algorithm>

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define pi acos(-1.0)



using namespace std;



typedef long long           ll      ;

typedef unsigned long long  ull     ;

typedef unsigned int        uint    ;

typedef unsigned char       uchar   ;



template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}

template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}



const double eps = 1e-7      ;

const int N = 210            ;

const int M = 1100011*2      ;

const ll P = 10000000097ll   ;

const int MAXN = 10900000    ;



int cnt[MAXN],a[MAXN];

bool check[MAXN];



void init_prime(){

    for(int i = 2; i < MAXN; ++i){ //素数打表

        if(!check[i]){

            for(int j = i; j < MAXN; j += i){

                check[j] = true;

                cnt[i] += a[j];

            }

        }

    }

}



int main(){

    std::ios::sync_with_stdio(false);

    int i, j, t, k, u, v, numCase = 0;

    int n, q, x, y;

    cin >> n;

    for(i = 1; i <= n; ++i){

            cin >> x;

            ++a[x];

    }

    init_prime();

    for(i = 2; i < MAXN; ++i){

        cnt[i] += cnt[i - 1];   //作累计和以节省查询时间

    }

    cin >> t;

    while(t--){

            cin >> x >> y;

            checkmin(x, 10000000);

            checkmin(y, 10000000);

            cout << cnt[y] - cnt[x - 1] << endl;

    }

    return 0;

}