[线段树][单调栈] BZOJ 4527 && CF 407E: K-D-Sequence

Solution

可以把同余的串一起考虑，然后把所有数除以d。变成公差为1的情况。
满足这样的条件

MAXL≤i≤RAi−MINL≤i≤RAi−R≤k−L

从右往左，用单调栈维护最大值最小值的位置，在线段树上修改上述值就好了。
感觉自己好傻逼。
注意不能有重复的数出现的串。

#include <bits/stdc++.h>using namespace std;const int N = 202020;const int M = 1000000000;const int INF = 1 << 30;inline char get(void) {    static char buf[100000], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, 100000, stdin);        if (S == T) return EOF;    }    return *S++;}template<typename T>inline void read(T &x) {    static char c; x = 0; int sgn = 0;    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';    if (sgn) x = -x;}int sta1[N], sta2[N];int ad[N << 2], mn[N << 2];int n, k, d, top1, top2;int cur, pos, ansl, ansr, lst;int a[N], b[N];map<int, int> usd;inline void PushDown(int o) {    if (ad[o]) {        ad[o << 1] += ad[o]; ad[o << 1 | 1] += ad[o];        mn[o << 1] += ad[o]; mn[o << 1 | 1] += ad[o];        ad[o] = 0;    }}inline void Add(int o, int l, int r, int L, int R, int x) {    if (l >= L && r <= R) {        ad[o] += x; mn[o] += x;        return;    }    int mid = (l + r) >> 1; PushDown(o);    if (L <= mid) Add(o << 1, l, mid, L, R, x);    if (R > mid) Add(o << 1 | 1, mid + 1, r, L, R, x);    mn[o] = min(mn[o << 1], mn[o << 1 | 1]);}inline void Find(int o, int l, int r, int L, int R, int x) {    if (l >= L && r <= R) {        if (mn[o] > x) return;        if (l == r) return (void)(pos = max(cur, l));        int mid = (l + r) >> 1; PushDown(o);        if (mn[o << 1 | 1] > x) return Find(o << 1, l, mid, L, R, x);        return Find(o << 1 | 1, mid + 1, r, L, R, x);    }    int mid = (l + r) >> 1; PushDown(o);    if (L <= mid) Find(o << 1, l, mid, L, R, x);    if (R > mid) Find(o << 1 | 1, mid + 1, r, L, R, x);}inline void Build(int o, int l, int r) {    ad[o] = 0; mn[o] = -r;    if (l == r) return;    int mid = (l + r) >> 1;    Build(o << 1, l, mid);    Build(o << 1 | 1, mid + 1, r);    mn[o] = min(mn[o << 1], mn[o << 1 | 1]);}inline void Solve(int L, int R) {    if (L == R) return;    sta1[top1 = 0] = sta2[top2 = 0] = R + 1;    usd.clear(); int l, r = R + 1;    for (l = R; l >= L; l--) {        if (usd.count(a[l])) r = min(r, usd[a[l]]);        while (top1 && a[l] > a[sta1[top1]]) {            Add(1, 1, n, sta1[top1], sta1[top1 - 1] - 1, -a[sta1[top1]]);            --top1;        }        sta1[++top1] = l;        Add(1, 1, n, sta1[top1], sta1[top1 - 1] - 1, a[sta1[top1]]);        while (top2 && a[l] < a[sta2[top2]]) {            Add(1, 1, n, sta2[top2], sta2[top2 - 1] - 1, a[sta2[top2]]);            --top2;        }        sta2[++top2] = l;        Add(1, 1, n, sta2[top2], sta2[top2 - 1] - 1, -a[sta2[top2]]);        pos = -1; Find(1, 1, n, l, r - 1, k - l);        if (~pos) {            if (pos - l > ansr - ansl)                ansl = l, ansr = pos;            if (pos - l == ansr - ansl && l < ansl)                ansl = l, ansr = pos;        }        usd[a[l]] = l;    }}int main(void) {    freopen("1.in", "r", stdin);    read(n); read(k); read(d);    if (d == 0) {        lst = 1; a[0] = INF; read(a[1]);        ansl = ansr = 1;        for (int i = 2; i <= n; i++) {            read(a[i]);            if (a[i] != a[lst]) {                if (i - 1 - lst > ansr - ansl)                    ansl = lst, ansr = i - 1;                lst = i;            }        }        printf("%d %d\n", ansl, ansr);        return 0;    }    for (int i = 1; i <= n; i++) {        read(a[i]); b[i] = (a[i] += M) % d;        a[i] /= d; ++a[i];    }    int j; Build(1, 1, n); ansl = ansr = 1;    for (int i = 1; i <= n; i = j) {        for (j = i + 1; j <= n; j++)            if (b[j] != b[i]) break;        Solve(i, j - 1);    }    printf("%d %d\n", ansl, ansr);    return 0;}