The Preliminary Contest for ICPC China Nanchang National Invitational

目录

• Contest Info
• Solutions
  • A. PERFECT NUMBER PROBLEM
  • D. Match Stick Game
  • G. tsy's number
  • H. Coloring Game'
  • I. Max answer
  • J. Distance on the tree
  • K. MORE XOR
  • M. Subsequence

Contest Info

---

[Practice Link](https://www.jisuanke.com/contest/2290?view=challenges)

Solved	A	B	C	D	E	F	G	H	I	J	K	L	M
8/13	O	-	-	O	-	-	O	O	O	O	O	-	O

• O 在比赛中通过
• Ø 赛后通过
• ! 尝试了但是失败了
• - 没有尝试

Solutions

---

A. PERFECT NUMBER PROBLEM

签到题。

#include <bits/stdc++.h>
using namespace std;

int main() {
	int a[] = {
		6, 28, 496, 8128, 33550336
	};
	for (int i = 0; i < 5; ++i) {
		printf("%d\n", a[i]);
	}
	return 0;
}

D. Match Stick Game

G. tsy's number

H. Coloring Game'

I. Max answer

题意：
定义一个区间\([l, r]\)的值为：

\[\begin{eqnarray*} f(l, r) = (max_{i = l}^r a_i) \cdot (\sum\limits_{i = l}^r a_i) \end{eqnarray*} \]

思路一：
单调栈求出当前点左边第一个比它小的位置，当前点右边第一个比它小的位置。
然后就算出管辖范围，然后线段树维护一下最大最小区间前后缀即可。
代码一：

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 500010
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3f
int n, a[N];
ll sum[N];
int f[N], g[N];
int Sta[N], top;

struct SEG {
	struct node {
		ll Max, Min;
		node () {
			Min = INFLL;
			Max = -INFLL;
		}
		node (ll Max, ll Min) : Max(Max), Min(Min) {}
		node operator + (const node &other) const {
			node res = node();
			res.Max = max(Max, other.Max);
			res.Min = min(Min, other.Min);
			return res;
		}
	}t[N << 2], res;
	void build(int id, int l, int r) {
		if (l == r) {
			t[id] = node(sum[l], sum[l]);
			return;
		}
		int mid = (l + r) >> 1;
		build(id << 1, l, mid);
		build(id << 1 | 1, mid + 1, r);
		t[id] = t[id << 1] + t[id << 1 | 1];
	}
	void query(int id, int l, int r, int ql, int qr) {
		if (ql > qr) {
			return;
		}
		if (l >= ql && r <= qr) {
			res = res + t[id];
			return;
		}
		int mid = (l + r) >> 1;
		if (ql <= mid) query(id << 1, l, mid, ql, qr);
		if (qr > mid) query(id << 1 | 1, mid + 1, r, ql, qr);
	}
}seg;

int main() {
	while (scanf("%d", &n) != EOF) {
		for (int i = 1; i <= n; ++i) {
			scanf("%d", a + i);
		}
		for (int i = 1; i <= n; ++i) {
			sum[i] = sum[i - 1] + a[i];
		}
		seg.build(1, 1, n);
		ll res = 0;

		a[0] = a[n + 1] = -INF;
		top = 0;
		Sta[++top] = 0;
		for (int i = 1; i <= n; ++i) {
			while (a[i] <= a[Sta[top]]) {
				--top;
			}
			f[i] = Sta[top];
			Sta[++top] = i;
		}
		
		top = 0;
		Sta[++top] = n + 1;
		for (int i = n; i >= 1; --i) {
			while (a[i] <= a[Sta[top]]) {
				--top;
			}
			g[i] = Sta[top];
			Sta[++top] = i;
		}
	//	for (int i = 1; i <= n; ++i) {
	//		printf("%d %d %d\n", i, f[i], g[i]);
	//	}
		for (int i = 1; i <= n; ++i) {
			if (a[i] == 0) {
				continue;
			} else if (a[i] < 0) {
				seg.res = SEG::node();
				ll x = 0, y = 0;	
				seg.query(1, 1, n, f[i], i);
				if (f[i] == 0) {
					x = max(x, seg.res.Max);
				} else {
					x = seg.res.Max;
				}
				seg.res = SEG::node();
				seg.query(1, 1, n, i, g[i] - 1);
				y = seg.res.Min;
				res = max(res, (y - x) * a[i]);
			} else {
				seg.res = SEG::node();
				ll x = 0, y = 0;
				seg.query(1, 1, n, f[i], i);
				if (f[i] == 0) {
					x = min(x, seg.res.Min);
				} else {
					x = seg.res.Min;
				}
				seg.res = SEG::node();
				seg.query(1, 1, n, i, g[i] - 1);
				y = seg.res.Max;
				res = max(res, (y - x) * a[i]);
			}
		}
			
		printf("%lld\n", res);
	}
	return 0;
}

思路二：
建出笛卡尔树，然后就确定了区间最小值，再考虑中序遍历是原序列。
那么左右子树分别维护区间和、区间最大最小前后缀，然后向上统计答案并合并

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 500010
#define INF 0x3f3f3f3f
int n, a[N];
ll res;

struct Cartesian_Tree {
	struct node {
		int id, val, fa;
		// 0 前缀最值
		// 1 后缀最值
		ll Min[2], Max[2], sum;
		int son[2];
		node() {}
		node (int id, int val, int fa) : id(id), val(val), fa(fa) {
			memset(son, 0, sizeof son);
			memset(Min, 0, sizeof Min);
			memset(Max, 0, sizeof Max);
			sum = 0;
		}
		bool operator < (const node &other) const {
			return id < other.id;
		}
		
	}t[N];
	int root;
	void init() {
		t[0] = node(0, -INF, 0);
	}
	void build(int n, int *a) {
		for (int i = 1; i <= n; ++i) {
			t[i] = node(i, a[i], 0);
		}
		for (int i = 1; i <= n; ++i) {
			int k = i - 1;

			while (t[k].val > t[i].val) {
				k = t[k].fa;
			}

			t[i].son[0] = t[k].son[1];
			t[k].son[1] = i;
			t[i].fa = k;
			t[t[i].son[0]].fa = i;
		}
		root = t[0].son[1]; 
	}
	void DFS(int u) {
		if (!u) return;
		int ls = t[u].son[0], rs = t[u].son[1];
		DFS(ls); DFS(rs);
		res = max(res, t[u].val * (t[u].val + t[ls].Min[1] + t[rs].Min[0]));
		res = max(res, t[u].val * (t[u].val + t[ls].Max[1] + t[rs].Max[0]));
		t[u].sum = t[ls].sum + t[rs].sum + t[u].val;  
		t[u].Min[0] = min(t[ls].Min[0], t[ls].sum + t[u].val + t[rs].Min[0]);
		t[u].Min[1] = min(t[rs].Min[1], t[rs].sum + t[u].val + t[ls].Min[1]);
		t[u].Max[0] = max(t[ls].Max[0], t[ls].sum + t[u].val + t[rs].Max[0]);
		t[u].Max[1] = max(t[rs].Max[1], t[rs].sum + t[u].val + t[ls].Max[1]);
	}
}CT;

int main() {
	while (scanf("%d", &n) != EOF) {
		for (int i = 1; i <= n; ++i) {
			scanf("%d", a + i);
		}
		res = -1e18;
		CT.init();		
		CT.build(n, a);
		CT.DFS(CT.root);
		printf("%lld\n", res);
	}
	return 0;
}

J. Distance on the tree

K. MORE XOR

M. Subsequence

题意：
给出串\(S\)，以及若干串\(T_i\)，每次询问\(T_i\)是否是\(S\)的一个子序列。

思路：
建出序列自动机，暴力跑即可。
时间复杂度：\(\mathcal{O}(26|S| + \sum T_i)\)

代码：

#include <bits/stdc++.h>
using namespace std;

#define N 100010
int n, m, q;
char s[N], t[N];
int T[N][30], nx[30];

int main() {
	while (scanf("%s", s + 1) != EOF) {
		n = strlen(s + 1);
		for (int i = 0; i < 30; ++i) nx[i] = n + 1;
		for (int i = n; i >= 0; --i) {
			for (int j = 0; j < 26; ++j) {
				T[i][j] = nx[j];	
			}
			if (i) {
				nx[s[i] - 'a'] = i;
			}
		}
		scanf("%d", &q);
		while (q--) {
			scanf("%s", t + 1);
			m = strlen(t + 1);
			int it = 0;
			for (int i = 1; i <= m; ++i) {
				it = T[it][t[i] - 'a'];	
				if (it == n + 1) break;
			}
			puts(it == n + 1 ? "NO" : "YES");
		}
	}
	return 0;
}