（SPOJ2）Prime Generator

最新推荐文章于 2021-02-12 14:47:40 发布

转载最新推荐文章于 2021-02-12 14:47:40 发布 · 70 阅读

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原文链接：http://www.cnblogs.com/cpoint/archive/2013/04/14/3021253.html

本文介绍了一个用于生成指定范围内所有素数的算法实现。输入包括测试案例数量及每组案例的两个整数边界值，输出为边界内的所有素数。算法采用筛选法去除合数，并针对大数据量进行了优化。

Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers!

Input

The input begins with the number t of test cases in a single line (t<=10). In each of the next t lines there are two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.

Output

For every test case print all prime numbers p such that m <= p <= n, one number per line, test cases separated by an empty line.

Example

Input:
2
1 10
3 5

Output:
2
3
5
7

3
5

Warning: large Input/Output data, be careful with certain languages (though most should be OK if the algorithm is well designed)

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>

using namespace std;

int main()
{
    int t;
    scanf("%d\n", &t);
    while(t--)
    {
        int n,m;
        scanf("%d %d\n", &n, &m);
        int * primes = new int[m-n+1];  
        for(int i=0;i<m-n+1;++i)
        primes[i] = 0;      //循环全部赋值为0 

        for(int p=2;p*p<=m;++p)
        {
            int less = n / p;
            less *= p;    // first number <= N && p divides N

            for(int j=less;j<=m;j+=p) 
             if(j != p && j >= n)
              primes[j - n] = 1;
        }

        for(int i=0;i<m-n+1;++i)
        {
            if(primes[i] == 0 && n+i != 1) //输出除去数字"1" 
            printf("%d\n",n+i);
        }

        if(casen)
          printf("\n");

        delete [] primes;
    }
}

转载于:https://www.cnblogs.com/cpoint/archive/2013/04/14/3021253.html