Acwing-196-质数距离(素数区间筛法)

链接:

https://www.acwing.com/problem/content/198/

题意:

给定两个整数L和U，你需要在闭区间[L,U]内找到距离最接近的两个相邻质数C1和C2（即C2-C1是最小的），如果存在相同距离的其他相邻质数对，则输出第一对。

同时，你还需要找到距离最远的两个相邻质数D1和D2（即D1-D2是最大的），如果存在相同距离的其他相邻质数对，则输出第一对。

思路:

筛除,l-r中的素数, 使用区间筛法, 先筛出1-sqrt(r), 再筛l-r.
记录l-r的素数,挨个判断.

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int MAXN = 1e6+10;
int IsPri1[MAXN], IsPri2[MAXN];
LL Pri[MAXN], Res[MAXN];
LL l, r;
int pos;

void Euler()
{
    memset(IsPri1, 0, sizeof(IsPri1));
    memset(IsPri2, 0, sizeof(IsPri2));
    pos = 0;
    for (int i = 2;i <= sqrt(r);i++)
    {
        if (IsPri1[i] == 0)
            Pri[++pos] = i;
        for (int j = 1;j <= pos && Pri[j]*i <= sqrt(r);j++)
        {
            IsPri1[Pri[j]*i] = 1;
            if (i%Pri[j] == 0)
                break;
        }
    }
    for (int i = 1;i <= pos;i++)
    {
        for (LL j = max(2LL, (0LL+l+Pri[i]-1)/Pri[i])*Pri[i];j <= r;j += Pri[i])
            IsPri2[j-l+1] = 1;
    }
}

int main()
{
    while (~scanf("%lld%lld", &l, &r))
    {
        Euler();
        int cnt = 0;
        int sl, sr, bl, br;
        int sv = 1e9, bv = 0;
        if (l == 1)
            IsPri2[l] = 1;
        for (int i = 1;i <= r-l+1;i++)
        {
            if (IsPri2[i] == 0)
                Res[++cnt] = i+l-1;
        }
        if (cnt < 2)
            printf("There are no adjacent primes.
");
        else
        {
            for (int i = 1;i < cnt;i++)
            {
                if (Res[i+1]-Res[i] < sv)
                {
                    sv = Res[i+1]-Res[i];
                    sl = Res[i], sr = Res[i+1];
                }
                if (Res[i+1]-Res[i] > bv)
                {
                    bv = Res[i+1]-Res[i];
                    bl = Res[i], br = Res[i+1];
                }
            }
            printf("%d,%d are closest, %d,%d are most distant.
", sl, sr, bl, br);
        }
    }

    return 0;
}