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质数筛 nbut1454 Lux

2015-08-25 22:36 337 查看
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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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