Inversion

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HDU 6098
这里写图片描述

线段树加优化
一开始妙想的就是线段树，时间是nlognlogn的，就是很无脑地计算每一段区间的最大值然后算出每一个Bi，但是超时了。在这个基础上加优化，注意到每一个合数，他的答案就是其每一个因数的答案的最大值，画一下就能看出来，加上这个就过了。

#include <bits/stdc++.h>using namespace std;const int maxn = 100010;const int maxm = 100000;int f[5*maxn], n, x, a[maxn], p[maxn];bool flag[maxn];void getPrime() {     memset(p, 0, sizeof(p));     memset(flag, 0, sizeof(flag));     for (int i = 2; i <= maxm; i++) {         if (!flag[i]) p[++p[0]] = i;         for (int j = 1; j <= p[0] && p[j] <= maxm/i; j++) {             flag[p[j]*i] = true;             if (i%p[j] == 0) break;         }     }}void insert(int x, int l, int r, int t, int v) {    if (l == r) {        f[x] = v;        return;    }    int mid = (l+r)/2;    if (t <= mid) insert(x*2, l, mid, t, v);    else insert(x*2+1, mid+1, r, t, v);    f[x] = max(f[x*2], f[x*2+1]);}int find(int x, int l, int r, int ll, int rr) {    if (l == ll && r == rr) {        return f[x];    }    int mid = (l+r)/2;    int t = 0;    if (rr <= mid) t = find(x*2, l, mid, ll ,rr);    else if (ll > mid) t = find(x*2+1, mid+1, r, ll, rr);    else {        t = find(x*2, l, mid, ll, mid);        t = max(t, find(x*2+1, mid+1, r, mid+1, rr));    }    return t;}int main() {    int T;    scanf("%d", &T);    while (T--) {        memset(f, 0, sizeof(f));        memset(a, 0, sizeof(a));        scanf("%d", &n);        for (int i = 1; i <= n; i++) {            scanf("%d", &x);            insert(1, 1, n, i, x);        }        for (int i = 2; i <= n; i++) {            if (!flag[i]) {                for (int j = 1; j <= n/i; j++) {                    a[i] = max(a[i], find(1, 1, n, (j-1)*i+1, j*i-1));                }                if (n/i*i < n) a[i] = max(a[i], find(1, 1, n, n/i*i+1, n));            } else {                for (int j = 2; j*j <= i; j++) if (i%j == 0) {                    a[i] = max(a[i], a[j]);                    a[i] = max(a[i], a[i/j]);                }            }        }        for (int i = 2; i < n; i++) printf("%d ", a[i]);        printf("%d\n", a[n]);    }}

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