poj 2689 Prime Distance

Prime Distance

Time Limit: 1000MS  Memory Limit: 65536K

Total Submissions: 29401  Accepted: 7556

Description

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 17 14 17 

Sample Output

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

题意：在区间[L,U]中找出一对距离最近的相邻素数和一对距离最远的相邻素数，距离相同时，取第一对。

题解：区间素数筛。由于数据范围太大，无法用数组维护素数。用区间[1，]的素数去更新区间[L，U]的素数。(用已知小范围素数推未知大范围素数)。区间间差不超多1000000，利用下标偏移,区间[0,U-L]维护区间[L,U]。

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int inf=0x3f3f3f3f;
int prime[5000],top;
bool flag[50005]={1,1},isp[1000005];
void init();
void solve(int L,int R);
int main()
{
  init();
  int L,R,l1,r1,l2,r2,i,j,mmin,mmax;
  while(~scanf("%d%d",&L,&R))
  {
    if(L==1) L=2; //重要
    {
      fill(isp,isp+1000005,1);
      mmin=inf;mmax=-inf;
    }
    solve(L,R);
    for(i=0;i<=R-L;i++)
    {
      if(!isp[i]) continue;
      for(j=i+1;j<=R-L;j++)
       if(isp[j]) 
        {
          if(mmax<(j-i)) mmax=j-i,l1=i+L,r1=j+L;
          if(mmin>(j-i)) mmin=j-i,l2=i+L,r2=j+L;
          break;
        }
    }
    if(mmax==-inf) printf("There are no adjacent primes.
");
    else printf("%d,%d are closest, %d,%d are most distant.
",l2,r2,l1,r1);
  }
  return 0;
}
void init()
{
  int i,j;
  for(i=2;i<50005;i++)
  {
   if(!flag[i]) prime[top++]=i;
   for(j=0;j<top&&i*prime[j]<50005;j++)
   {
     flag[i*prime[j]]=1;
     if(i%prime[j]==0) break;
   }
  }
}
void solve(int L,int R)
{
  long long i,l;
  for(i=2;i*i<=R;i++)
  {
    if(flag[i]) continue;
    for(l=max(2LL,((i+L-1)/i))*i;l<=R;l+=i) isp[l-L]=0;
  }
}