poj2689 素数距离（埃氏筛法分块筛+暴力枚举）

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#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#include<stack>#include<queue>#include<vector>#include<set>#include<map>#include<string>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;}
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