[Agc036D]Do Not Duplicate_链表_贪心_数论

Do Not Duplicate

题目链接：https://atcoder.jp/contests/agc036/tasks/agc036_b

---

题解：

首先最后肯定至多只有$n$个数。

我们想处理出来每个点下一个一样的数的下一个数。

有点绕口....
处理出来了之后，暴力找环然后暴力跳就好。

代码：

#include <bits/stdc++.h>

#define N 200010 

using namespace std;

typedef long long ll;

int a[N], pre[N], nxt[N], val[N];

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
	int x = 0;
	char c = nc();
	while (c < 48) {
		c = nc();
	}
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
	}
	return x;
}

ll rd2() {
	ll x = 0;
	char c = nc();
	while (c < 48) {
		c = nc();
	}
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
	}
	return x;
}

int First[N];

int main() {
	int n = rd();
	ll k = rd2();
	for (int i = 1; i <= n; i ++ ) {
		a[i] = rd();
	}
	if (n == 1) {
		if (k & 1) {
			printf("%d
", a[1]);
		}
		return 0;
	}
	for (int i = n; i; i -- ) {
		First[a[i]] = i;
	}
	for (int i = 1; i <= n; i ++ ) {
		if (pre[a[i]]) {
			nxt[pre[a[i]]] = (i + 1) % n;
			if (!nxt[pre[a[i]]]) {
				nxt[pre[a[i]]] = n;
			}
		}
		pre[a[i]] = i;
	}
	for (int i = 1; i <= n; i ++ ) {
		if (!nxt[i]) {
			if (First[a[i]] != i) {
				nxt[i] = First[a[i]] + 1;
			}
			else {
				if (i == n) {
					nxt[i] = 1;
				}
				else {
					nxt[i] = i + 1;
				}
			}
		}
	}
	for (int i = 1; i <= n; i ++ ) {
		if (nxt[i] >= i + 2) {
			val[i] = nxt[i] - i;
		}
		else {
			val[i] = nxt[i] + n - i;
		}
	}
	if (nxt[n] == 1) {
		val[n] = n + 1;
	}
	ll mdl = 0;
	int now = 1;
	while (1) {
		mdl += val[now];
		now = nxt[now];
		if (now == 1) {
			break;
		}
	}
	ll pre = mdl / n;
	// cout << pre << endl ;
	k %= pre;
	if (!k) {
		return 0;
	}
	// puts("Fuck");
	int id = 1;
	now = 1;
	while (1) {
		if (nxt[now] >= now + 2) {
			now = nxt[now];
			continue;
		}
		if (id == k) {
			if (nxt[now] == 1 && now != n) {
				return 0;
			}
			printf("%d ", a[now]);
			if (now == n) {
				return 0;
			}
			now ++ ;
		}
		else {
			id ++ ;
			now = nxt[now];
		}
	}
	return 0;
}

小结：想题的时候，多想一些特殊情况。比如边界值啊，极值啊这种。