POJ 2689 Prime Distance(区间素数筛)

题目链接

题目大意

  求一个区间内相差最小的一对相邻素数，和相差最大的一对相邻素数。

具体实现

  注意越界。

代码

const int maxn = 2e6+10;
ll u[maxn], p[maxn], tmp[maxn], kase;
bool isp[maxn];
void pri() {
    for (int i = 2; i<maxn; ++i) {
        if (!u[i]) u[i] = p[++kase] = i;
        for (int j = 1; i*p[j]<maxn; ++j) {
            u[i*p[j]] = p[j];
            if (!(i%p[j])) break;
        }
    }
}
int main(void) {
    pri();
    ll l, r;
    while(~scanf("%lld%lld", &l, &r)) {
        zero(isp);
        for (int i = 1; 1LL*p[i]*p[i]<=r; ++i) //区间筛选
            for (ll j = l/p[i]*p[i]; j<=r; j+=p[i])
                if (j>p[i] && j>=l) isp[j-l] = true;
        int tot = 0;
        for (ll i = l; i<=r; ++i) //将素数存入一个数组中
            if (!isp[i-l] && i!=1) tmp[tot++] = i;
        ll maxx = -1, minn = INF;
        P ans1 = P(0,0), ans2 = P(0,0);
        for (int i = 1; i<tot; ++i) {
            if (maxx < tmp[i]-tmp[i-1]) {
                ans1 = P(tmp[i-1], tmp[i]);
                maxx = tmp[i]-tmp[i-1];
            }
            if (minn > tmp[i]-tmp[i-1]) {
                ans2 = P(tmp[i-1],tmp[i]);
                minn = tmp[i]-tmp[i-1];
            }
        }
        if (!ans1.first) printf("There are no adjacent primes.
");
        else printf("%d,%d are closest, %d,%d are most distant.
", ans2.first,ans2.second,ans1.first,ans1.second);
    }
    return 0;
}