poj 2689 （素数二次筛选）

 

Sample Input

2 17
14 17

Sample Output

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

找出给定范围内，距离最远和最近的素数。（不停超时 - -）

给的上界很大，所以全处理肯定不行。 先处理sqrt（2147483647）。

然后再在l 与 r之间筛选素数。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define N 10100
typedef long long ll;
using namespace std;
int n,m;
int f[1000500];
int q[1000500];
int t[100050];
int tot;
void prim()
{
    ll i,j;
    for(i = 2; i <= 100050; i++)
        q[i] = 1;
    for(i= 2,tot = 0; i <= 100050; i++)
    {
        if(q[i])
        {
            t[tot++] = i;
            for(j = i*2; j <= 100050; j+=i)
                q[j] = 0;
        }
    }
}


int main()
{
    int l,r;
    while(scanf("%d%d",&l,&r)!= EOF)
    {
        prim();
        if(l == 1)
            l = 2;

        memset(f,0,sizeof(f));
        for(int i =0; i < tot; i++)
        {
            int a = (l-1)/t[i]+1;
            int b = r /t[i];
            for(int j = a; j <= b; j++)     //类似上面prim（）的方法
                if(j > 1)
                    f[j*t[i]-l] = 1;
        }
        int Max = -1;
        int Min = 1000000000;
        int tmp = -1;
        int maxl,maxr,minl,minr;

        for(int i = 0; i <= r-l; i++)
        {
            if(!f[i])
            {
                if(tmp>=0 && i - tmp> Max)
                {
                    Max = i-tmp;
                    maxl = tmp+l;
                    maxr = i+l;
                }
                if(tmp>=0 && i - tmp< Min)
                {
                    Min = i-tmp;
                    minl = tmp+l;
                    minr = i+l;
                }
                tmp = i;
            }
        }
        if(Max == -1)
            printf("There are no adjacent primes.\n");
        else
        {
            printf("%d,%d are closest, %d,%d are most distant.\n",minl,minr,maxl,maxr);
        }
    }
    return 0;
}