POJ2689 Prime Distance

　　原题传送：http://poj.org/problem?id=2689

　　求相连素数最小距离和最大距离。

　　由于n很大，但是U-L较小，所以筛素数时只筛到sqrt(n)就可以了，利用这个已经筛出的素数表把[U,L]之间的素数筛出来。

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#define LL long long
#define N 50000
#define M 1000001

LL base[N], res[M];
LL l, u, k;
bool isprime[N], isres[M];
LL c1, c2, d1, d2;

void init()  // 筛素数
{
    LL i, j;
    memset(isprime, false, sizeof isprime);
    for(i = 2, k = 0; i < N; i ++)
    {
        if(!isprime[i])
        {
            base[k ++] = i;
            for(j = i << 1; j < N; j += i)
                isprime[j] = true;
        }
    }
}

void solve()
{
    LL i, j, cnt;
    memset(isres, false, sizeof isres);
    for(i = 0; i < k; i ++)
    {
        LL tmp = l / base[i];      // 快速定位
        while(tmp * base[i] < l || tmp <= 1)
            ++ tmp;
        for(j = tmp * base[i]; j <= u; j += base[i])
            isres[j - l] = true;
    }
    if(l == 1)
        isres[0] = true;
    for(cnt = 0, i = 0; i <= u - l; i ++)
    {
        if(!isres[i])
            res[cnt ++] = i + l;
    }
    if(cnt <= 1)
        puts("There are no adjacent primes.");
    else
    {
        LL max = 0, min = 0x7fffffff;
        for(i = 1; i < cnt; i ++)
        {
            if(res[i] - res[i - 1] < min)
            {
                min = res[i] - res[i - 1];
                c1 = res[i - 1];
                c2 = res[i];
            }
            if(res[i] - res[i - 1] > max)
            {
                max = res[i] - res[i - 1];
                d1 = res[i - 1];
                d2 = res[i];
            }
        }
        printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.\n", c1, c2, d1, d2);
    }
}

int main()
{
    init();
    while(scanf("%lld%lld", &l, &u) != EOF)
        solve();
    return 0;
}