poj2689 Prime Distance

考虑用筛法排除区间内所有合数，对余下的数（素数）更新答案。

http://poj.org/problem?id=2689

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef __int64 LL;
const int maxn = 5e4;
const int maxm = 1e6 + 10;
LL prime[6000], k;
LL f[maxm], t;
bool vis[maxm];
bool np[maxm];
LL a, b;

void init(){
    memset(vis, 0, sizeof vis);
    k = 0;
    for(int i = 2; i < maxn; i++){
        if(!vis[i]) prime[k++] = (LL)i;
        for(int j = i; j < maxn; j += i) vis[j] = 1;
    }
}

void solve(){
    //[a, b] [0, b - a]
    if(a == 1) ++a;
    memset(np, 0, sizeof np);
    for(LL i = 0; i < k; i++){
        LL tem = a % prime[i];
        LL j0 = tem == 0 ? a : a + prime[i] - tem;
        tem = b % prime[i];
        LL j1 = tem == 0 ? b : b - tem;
        if(j0 == prime[i]) j0 += prime[i];
        for(LL j = j0; j <= j1; j += prime[i]) np[j - a] = 1;
    }
    LL maxi = -1, mini = 0x3f3f3f3f;
    LL a1, b1;
    t = 0;
    for(int i = 0; i <= b - a; i++) if(!np[i]) f[t++] = i;
    for(int i = 0; i < t - 1; i++){
        int d = f[i + 1] - f[i];
        if(d > maxi) a1 = f[i] + a, maxi = d;
        if(d < mini) b1 = f[i] + a, mini = d;
    }
    if(maxi != -1){
        printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.
", b1, b1 + mini, a1, a1 + maxi);
    }else puts("There are no adjacent primes.");
}

int main(){
    //freopen("in.txt", "r", stdin);
    init();
    while(~scanf("%I64d%I64d", &a, &b)) solve();
    return 0;
}