【筛选法】 POJ 2689 Prime Distance

筛选法好题。。。用筛选法打出2-2的16次方的素数表，然后再对l到r做筛选法。。。不过要注意特殊处理1。。

#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>#include <cmath>#include <time.h>#define maxn 1000005#define maxm 3000005#define eps 1e-10#define mod 998244353#define INF 999999999#define lowbit(x) (x&(-x))#define mp mark_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R//#define debug(x) printf("AA x = %d BB\n", x);//#pragma comment (linker,"/STACK:102400000,102400000")typedef long long LL;//typedef int LL;using namespace std;LL powmod(LL _a, LL _b){LL _res=1,_base=_a;while(_b){if(_b%2)_res=_res*_base%mod;_base=_base*_base%mod;_b/=2;}return _res;}void scanf(LL &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}// headint phi[maxn], prime[maxn], p[maxn];int cnt, l, r;void init(void){cnt = 0;for(int i = 2; i <= 65536; i++)if(!prime[i]) {phi[++cnt] = i;for(int j = i+i; j <= 65536; j+=i)prime[j] = 1;}}void work(void){int a, res, ans_mi, ans_mx, mi, mx;memset(prime, 0, sizeof prime);for(int i = 1; i <= cnt; i++) {if(l%phi[i] == 0) a = 0;else a = phi[i] - l%phi[i];if(l <= phi[i]) for(int j = a+phi[i]; j <= r-l; j+=phi[i]) prime[j] = 1;else for(int j = a; j <= r-l; j+=phi[i]) prime[j] = 1;}res = 0;for(int i = 0; i <= r-l; i++) if(!prime[i] && i+l != 1) p[++res] = i+l;if(res <= 1) printf("There are no adjacent primes.\n");else {mx = 0, mi = INF;for(int i = 1; i < res; i++) {if(mx < p[i+1] - p[i]) mx = p[i+1] - p[i], ans_mx = i;if(mi > p[i+1] - p[i]) mi = p[i+1] - p[i], ans_mi = i;}printf("%d,%d are closest, %d,%d are most distant.\n", p[ans_mi], p[ans_mi+1], p[ans_mx], p[ans_mx+1]);}}int main(void){init();while(scanf("%d%d", &l, &r)!=EOF) work();return 0;}