POJ2689 Prime Distance

题意：输入[l,r]代表，问区间里面最近和最远的两个素数（0<l<r<int_max, r-l<100000)

题解：素数筛，一个数是合数，那么它存在小于sqrt(n)的素因子，找到所有小于sqrt(int_max)的素数，注意数据：1 2

#include <iostream>
#include <string.h>
#include <stdio.h>
#define N 1001000
#define ll long long
using namespace std;
bool isprime[N], isp[N];
long long prime[N], num, ans[N];
void doprime(long long n){
    long long i,j;
    num = 0;
    memset(isprime, true,sizeof(isprime));
    isprime[1] = 0;
    for(i=2;i<=n;i++){
        if(isprime[i]){
            prime[num++] = i;
            for(j=i*i;j<=n;j+=i) isprime[j] = false;
        }
    }
}
void doprime2(long long l,long long r){
    memset(isp, true, sizeof(isp));
    for(long long i=0;i<num;i++){
        long long t = l/prime[i];
        while(prime[i]*t<l||t<=1) t++;
        for(long long j=t*prime[i];j<=r;j+=prime[i])
            isp[j-l] = 0;
    }
    if(l==1) isp[0] = 0;
}
int main(){
    long long ansl,ansr, l, r, ansll, ansrr;
    doprime(50000);
    while(cin>>l>>r){
        doprime2(l,r);
        long long t = 0;
        for(long long i=l;i<=r;i++)
            if(isp[i-l] == 1)
                ans[t++] = i;
        if(t < 2) printf("There are no adjacent primes.
");
        else{
            long long mi = 10000000, ma = -10000000;
            for(long long i=0;i<t-1;i++){
                if(mi>ans[i+1]-ans[i]){
                    ansl = ans[i];
                    ansr = ans[i+1];
                    mi = ansr-ansl;
                }
                if(ma < ans[i+1]-ans[i]){
                    ansll = ans[i];
                    ansrr = ans[i+1];
                    ma = ans[i+1]-ans[i];
                }
            }
            printf("%lld,%lld are closest, %lld,%lld are most distant.
", ansl, ansr, ansll, ansrr);
        }
    }
}