质数筛 nbut1454 Lux

传送门：点击打开链接

题意：给出区间[L,R],R-L<=1e6,求区间中相邻质数之差的最大值和最小值，如果不存在两个质数，就输出-1

思路：筛[L,R]区间中的全部质数，然后再遍历一遍

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;

const int MX = 1e6 + 5;
const int INF = 0x3f3f3f3f;

bool sign[MX];
int subprim[MX], prim[MX], totleprm = 0;

void getprim(int n) {
    memset(sign,0,sizeof(sign));

    sign[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(sign[i]) continue;
        prim[totleprm++] = i;

        if((LL)i * i > n) continue;
        for(int j = i * i; j <= n; j += i) {
            sign[j] = 1;
        }
    }
}

int getSubPrim(int a, int b) {
    int totl = 0, i;LL j;
    memset(sign, true, sizeof(sign));

    if(a < 2) a = 2;
    LL l = b - a + 1;
    for(i = 0; i < totleprm; i++) {
        if((j = prim[i] * (a / prim[i])) < a) j += prim[i];
        if(j < prim[i]*prim[i]) j = prim[i] * prim[i];
        for(; j <= b; j += prim[i]) sign[j - a] = false;
    }
    for(i = 0; i < l; i++) if(sign[i]) {
            subprim[totl++] = a + i;
        }
    return totl;
}

int main() {

    getprim(100000);
    int L, R;//FIN;
    while(~scanf("%d%d", &L, &R)) {
        int MinL = 0, MinR = INF, MaxL = 0, MaxR = 0;
        int rear = getSubPrim(L, R);
        if(rear < 2) {
            printf("-1\n");
            continue;
        }

        for(int i = 0; i < rear - 1; i++) {
            if(MinR - MinL > subprim[i + 1] - subprim[i]) MinL = subprim[i], MinR = subprim[i + 1];
            if(MaxR - MaxL < subprim[i + 1] - subprim[i]) MaxL = subprim[i], MaxR = subprim[i + 1];
        }
        printf("%d %d %d %d\n", MinL, MinR, MaxL, MaxR);
    }
    return 0;
}

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