Spoj PRIME1 - Prime Generator

Posted 2020-11-20 lovewhy

题意翻译

求给定的两个数之间的素数

Translated by @kaiming

题目描述

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

输入输出格式

输入格式：

 

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.

 

输出格式：

 

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.

 

输入输出样例

输入样例#1： 复制

2
1 10
3 5

输出样例#1： 复制

2
3
5
7

3
5

说明

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

After cluster change, please consider PRINT as a more challenging problem.

 

//Pro: Spoj PRIME1 - Prime Generator
//求给定的两个数之间的素数 

#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;

const long long N=40000;

long long T;
long long cnt;
long long prime[N+1];
bool flag[N+1];

void init()
{
    flag[1]=1;
    long long tmp;
    for(long long i=2;i<=N;++i)
    {
        if(!flag[i])
            prime[++cnt]=i;
        for(long long j=1;j<=cnt&&(tmp=i*prime[j])<=N;++j)
        {
            flag[tmp]=1;
            if(i%prime[j]==0)
                break;
        }
    }
}


long long l,r;
bool Flag;
int main()
{
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&l,&r);
        for(;l<=r;++l)
        {
            if(l<=N)
            {
                if(!flag[l])
                {
                    printf("%d
",l);
                    continue;
                }
            }
            else
            {
                Flag=0;
                for(long long j=1;prime[j]<=sqrt(l)&& !Flag && j<=cnt;++j)
                {
                    if(l%prime[j]==0)
                        Flag=1;
                }
                if(!Flag)
                    printf("%d
",l);
            }
        }
        puts("");
    }
    return 0;
}