2016春季练习——二次筛

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=50010;
const long long INF=0x7fffffff;
long long prime1
,np1;
bool isp[N*200];
void make_p1(){
long long i,j;
np1=0;
memset(isp,1,sizeof(isp));
isp[0]=isp[1]=0;
for(i=2;i<N;i++){
if(isp[i]){
np1++;
prime1[np1]=i;
for(j=i*i;j<N;j+=i)
isp[j]=0;
}
}
}
long long l,u;
long long prime2[1000100];
int np2;
void make_p2(){
long long i,j,b;
memset(isp,1,sizeof(isp));
for(i=1;i<=np1;i++){
b=l/prime1[i];
while(b*prime1[i]<l||b<=1)
b++;
for(j=b*prime1[i];j<=u;j+=prime1[i]){
if(j>=l)
isp[j-l]=0;
}
}
if(l==1)
isp[0]=0;
}
void solve(){
int i;
long long mina=INF,maxn=-INF;
long long minl,minr,maxl,maxr;
make_p2();
np2=0;
for(i=0;i<u;i++){
if(isp[i]){
np2++;
prime2[np2]=i+1;
}
}
if(np2<=l){
cout<<"There are no adjacent primes."<<endl;
return ;
}
else{
for(i=1;i<np2;i++){
if(prime2[i+1]-prime2[i]<mina){
mina=prime2[i+1]-prime2[i];
minl=prime2[i],minr=prime2[i+1];
}
if(prime2[i+1]-prime2[i]>maxn){
maxn=prime2[i+1]-prime2[i];
maxl=prime2[i],maxr=prime2[i+1];
}
}
printf("%lld,%lld are closest, %lld,%lld are most distant.\n",minl,minr,maxl,maxr);
}
}
int main(){
make_p1();
while(scanf("%lld%lld",&l,&u)!=EOF){
solve();
}
return 0;
}

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF=1e9;
const int maxn=5e4+10;
const int maxm=1e6+10;
int l,u,prime[maxn],vis[maxn],f[maxm],t;
void ini()//求出50000内的素数，int范围内的合数最小质因子必定小于2^16。
{
int i,j,k,m;
t=0;
m=(int)sqrt(maxn+0.5);
memset(vis,0,sizeof(vis));
for(i=2;i<=m;i++)
{
if(!vis[i])
{
for(j=i*i;j<maxn;j+=i)
vis[j]=1;
}
}
for(i=2;i<maxn;i++)if(!vis[i])prime[t++]=i;
}
int main()
{
ini();
while(cin>>l>>u)
{
if(l==1)l=2;//注意l=1会出问题。
int i,j,k,a,b;
memset(f,0,sizeof(f));
for(i=0;i<t;i++)
{
a=(l-1)/prime[i]+1;
b=u/prime[i];
for(j=a;j<=b;j++)if(j>1)f[j*prime[i]-l]=1;//[l,u]区间小于1e6，而l，u数值范围为int，所以偏移l们就能用数组存了
}
int p=-1,max_ans=-1,min_ans=INF,x1,y1,x2,y2;
for(i=0;i<=u-l;i++)//暴力枚举。。
{
if(f[i]==0)
{
if(p==-1){p=i;continue;}
if(max_ans<i-p){max_ans=i-p;x1=p+l;y1=i+l;}
if(min_ans>i-p){min_ans=i-p;x2=p+l;y2=i+l;}
p=i;
}
}
if(max_ans==-1)cout<<"There are no adjacent primes."<<endl;
else cout<<x2<<","<<y2<<" are closest, "<<x1<<","<<y1<<" are most distant."<<endl;
}
return 0;
}