poj2689 素数距离

题意: 给出一个区间 [l, r] 求其中相邻的距离最近和最远的素数对 . 其中 1 <= l <  r <= 2,147,483,647, r - l <= 1e6 .

这个题就直接用上篇文章讲到的筛法，筛两次即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll INFF=1e12;
const int INF=0x7f7f7f7f;
ll l,r;
int cnt;
int prim[50010];
bool vis[1000010];
ll prim2[1000010];
void init()
{
    cnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<=50000;i++)
    {
        if(!vis[i])
        {
            prim[cnt++]=i;
            for(ll j=i+i;j<=50000;j+=i)
                vis[j]=1;
                //printf("%d\n",i);
        }
    }
}
void init2()
{
    ll i,j,b;
    memset(vis,0,sizeof(vis));
    for(i=0;i<cnt;i++)
    {
        b=l/prim[i];
        while(b*prim[i]<l||b<=1)
            b++;
        for(j=b*prim[i];j<=r;j+=prim[i])
        {
            if(j>=l)
                vis[j-l]=1;
        }
    }
    if(l==1)
        vis[0]=1;
}
 
void solve()
{
    init2();
    int cntt=0;
    for(int i=0;i<=r-l;i++)
    {
        if(!vis[i])
        {
            prim2[cntt++]=i+l;
        }
    }
    ll minn=INFF,maxx=-INFF;
    ll maxl,maxr,minl,minr;
    if(cntt<=1)
    {
        printf("There are no adjacent primes.\n");
    }
    else
    {
        for(int i=0;i<cntt-1;i++)
        {
            if(prim2[i+1]-prim2[i]<minn)
            {
                minn=prim2[i+1]-prim2[i];
                minl=prim2[i],minr=prim2[i+1];
            }
            if(prim2[i+1]-prim2[i]>maxx)
            {
                maxx=prim2[i+1]-prim2[i];
                maxl=prim2[i],maxr=prim2[i+1];
            }
        }
        printf("%lld,%lld are closest, %lld,%lld are most distant.\n",minl,minr,maxl,maxr);
    }
}
int main()
{
    init();
    while(scanf("%lld%lld",&l,&r)!=EOF)
    {
        solve();
    }
    return 0;
}