素数——POJ-2689

题目链接

题目代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn=5e4+7;
const int maxm=1000007;
int prime[maxn];
bool check[maxm];
int cnt=0;
void Prime(){
    for(int i=2;i<maxn;i++)
    if(!check[i]){
        prime[cnt++]=i;
        for(int j=2;j*i<maxn;j++)
            check[j*i]=true;
    }
}
LL l,r;
int main(){
    Prime();
    while(~scanf("%lld%lld",&l,&r)){
        int flag=0;
        if(l==1)l=2;
        LL minl,minr,maxl,maxr,pre=-1;
        memset(check,false,sizeof(check));
        for(int i=0;i<cnt;i++){
            LL a=(l-1)/prime[i]+1,b=r/prime[i];
            for(LL j=a;j<=b;j++)
            if(j>1){
                check[j*prime[i]-l]=true;
            }
        }
        for(LL i=0;i<=r-l;i++){
            if(!check[i]&&!flag){
                flag++;
                minl=maxl=i;
            }
            else if(!check[i]){
                flag++;
                if(flag==2){
                    minr=maxr=i;
                }
                else{
                    if(i-pre<minr-minl)minr=i,minl=pre;
                    if(i-pre>maxr-maxl)maxr=i,maxl=pre;
                }
                pre=i;
            }
        }
        if(flag<2)printf("There are no adjacent primes.
");
        else printf("%lld,%lld are closest, %lld,%lld are most distant.
",minl+l,minr+l,maxl+l,maxr+l);
    }
    return 0;
}