poj 2689 区间素数筛

The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.
Your program is given 2 numbers: L and U (1<=L<
U<=2,147,483,647), and you are to find the two adjacent primes C1 and
C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the
minimum). If there are other pairs that are the same distance apart, use
the first pair. You are also to find the two adjacent primes D1 and D2
(L<=D1< D2<=U) where D1 and D2 are as distant from each other
as possible (again choosing the first pair if there is a tie).

Input

Each line of input will contain two positive integers, L and U,
with L < U. The difference between L and U will not exceed 1,000,000.

Output

For each L and U, the output will either be the statement that
there are no adjacent primes (because there are less than two primes
between the two given numbers) or a line giving the two pairs of
adjacent primes.

Sample Input

2 17
14 17

Sample Output

2,3 are closest, 7,11 are most distant.
There are no adjacent primes.

题意 ： 求一下所给区间的相邻两素数的最大差值和最小差值

bool prime[47000];
bool pt[eps];

int main() {
    ll a, b;

    memset(prime, true, sizeof(prime));
    for(int i = 2; i < 47000; i++){
        if (prime[i]){
            for(int j = i+i; j < 47000; j += i){
                prime[j] = false;
            }
        }
    }

    while(~scanf("%lld%lld", &a, &b)){
        memset(pt, true, sizeof(pt));

        for(ll i = 2; i*i <= b; i++){
            if (prime[i]){
                for(ll j = i*max(2LL, (a+i-1)/i); j <= b; j += i){
                    pt[j-a] = false;
                }
            }
        }
        ll aa = -1, bb = -1, cc = -1, dd = -1;
        ll last, pr;
        ll t1 = 9999999, t2 = -1;
        int sign = 1;
        for(ll i = 0; i <= b-a; i++){
            if (pt[i]){
                pr = i + a;
                if (pr == 1) continue;
                ll cha = pr - last;

                if (sign){
                    last = pr;
                    sign = 0;
                    continue;
                }
                if (cha < t1){
                    t1 = cha;
                    aa = last;
                    bb = pr;
                }
                if (cha > t2){
                    t2 = cha;
                    cc = last;
                    dd = pr;
                }
                last = pr;
            }

        }
        if (aa == -1){
            printf("There are no adjacent primes.\n");
        }
        else {
            printf("%lld,%lld are closest, %lld,%lld are most distant.\n", aa, bb, cc, dd);
        }
    }

    return 0;
}