joj1928

 1928: Prime Distance

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ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K9925Standard

The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.

Your program is given 2 numbers: L and U (1<=L<U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1<C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1<D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).

Input

Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.

Output

For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.

Sample Input

2 1714 17

Output for Sample Input

2,3 are closest, 7,11 are most distant.There are no adjacent primes.

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This problem is used for contest: 190 

这是一个让人非常蛋疼的一个题目，这个题目太阴了，原来题目中用红色标注的数字是int所能表示的最大值。因此在计算过程中

很可能会出现比这个值更大的数值，因此计算过程中必须用long long。。。。。

#include<iostream>
#include<stdio.h>
using namespace std;
bool prime1[50000+1];
bool prime2[1000000+1];
int prime3[1000000+1];
int main()
{
    for(int i=3;i<=50000;i++)
    {
        if(i%2==0)prime1[i]=0;
        else prime1[i]=1;
    }
    for(int i=3;i*i<=50000;i++)
    {
        if(prime1[i])
        {
            for(int j=i*i;j<=50000;j+=i)prime1[j]=0;
        }
    }
    prime1[2]=1;prime1[1]=prime1[0]=0;
    int l,c;
    while(scanf("%d%d",&l,&c)==2)
    {
        for(long long  i=l;i<=c;i++)//此处的i必须是longlong否则c+1就会超出int的范围
        {
            if(i%2==0)prime2[i-l]=0;
            else prime2[i-l]=1;
        }
        for(long long  i=3;i*i<=c;i+=2)//此处i必须是longlong型的否则就超出int的范围了。
        {
            if(prime1[i])
            {
                for(long long  j=(l/i>2?l/i:2)*i;j<=c;j+=i)
                {
                    if(j-l>=0)
                    prime2[j-l]=0;
                }
            }
        }
        if(l==2)prime2[0]=1;
        if(l==1)
        {
            prime2[0]=0;prime2[1]=1;
        }
        int count=0;
        for(int i=0;i<=c-l;i++)
        {
            if(prime2[i])
            {
                prime3[++count]=i+l;
            }
        }
        if(count<=1)cout<<"There are no adjacent primes."<<endl;
        else
        {
            int max=-1;int min=c;
            int hmax,tmax,hmin,tmin;
            for(int i=1;i<=count-1;i++)
            {
                if(prime3[i+1]-prime3[i]>max)
                {
                    max=prime3[i+1]-prime3[i];
                    hmax=prime3[i];tmax=prime3[i+1];
                }
                if(prime3[i+1]-prime3[i]<min)
                {
                    min=prime3[i+1]-prime3[i];
                    hmin=prime3[i];tmin=prime3[i+1];
                }
            }
            printf("%d,%d are closest, %d,%d are most distant.\n",hmin,tmin,hmax,tmax);
        }
    }
    return 0;
}