hdu 4630 No Pain No Game（线段树）

题目链接：hdu 4630 No Pain No Game

解题思路

P[x]表示因子x最近出现的为值，在考虑位置i时，如果有因子x，则P[x]~i之间即可以有因子x，用线段树维护区间最大值。然后将查询离线。

代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

#define lson(x) (x<<1)
#define rson(x) ((x<<1)|1)
const int maxn = 50000;

int lc[maxn<<2], rc[maxn<<2], s[maxn<<2];

void pushup(int u) { s[u] = max(s[lson(u)], s[rson(u)]); }

void build(int u, int l, int r) {
    lc[u] = l, rc[u] = r, s[u] = 0;

    if (l == r) return;
    int mid = (l + r) >> 1;
    build(lson(u), l, mid);
    build(rson(u), mid+1 , r);
    pushup(u);
}

void modify(int u, int p, int v) {
    if (lc[u] == rc[u]) {
        s[u] = max(s[u], v);
        return;
    }

    int mid = (lc[u] + rc[u]) >> 1;
    if (p <= mid) modify(lson(u), p, v);
    else modify(rson(u), p, v);
    pushup(u);
}

int query(int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return s[u];

    int mid = (lc[u] + rc[u]) >> 1, ret = 0;
    if (l <= mid) ret = max(ret, query(lson(u), l, r));
    if (r > mid) ret = max(ret, query(rson(u), l, r));
    return ret;
}

struct State {
    int l, r, id;
    State(int l = 0, int r = 0, int id = 0):l(l), r(r), id(id) {}
    bool operator < (const State& u) const { return r < u.r; }
}Q[maxn + 5];

int N, M, A[maxn + 5], P[maxn + 5], ans[maxn + 5];
vector<int> G[maxn + 5];

void init () {
    scanf("%d", &N);
    memset(P, 0, sizeof(P));
    for (int i = 1; i <= N; i++) scanf("%d", &A[i]);

    scanf("%d", &M);
    for (int i = 0; i < M; i++) {
        scanf("%d%d", &Q[i].l, &Q[i].r);
        Q[i].id = i;
    }
    sort(Q, Q + M);
    build(1, 0, N);
}

int main () {
    for (int i = 1; i <= maxn; i++)
        for (int j = i; j <= maxn; j += i)
            G[j].push_back(i);

    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();

        int p = 0;
        for (int i = 1; i <= N; i++) {
            int u = A[i];

            for (int j = 0; j < G[u].size(); j++) {
                int v = G[u][j];
                modify(1, P[v], v);
                P[v] = i;
            }
            while (p < M && Q[p].r == i) {
                ans[Q[p].id] = query(1, Q[p].l, Q[p].r);
                p++;
            }
        }
        for (int i = 0; i < M; i++)
            printf("%d\n", ans[i]);
    }
    return 0;
}