2019杭电多校第五场

2019杭电多校第五场

1002. three arrays

upsloved

你有两个长为\(n\)的序列\(a, b\)，你可以任意打乱这两个序列，使得序列\(c\)字典序最小\((c_i = a_i\, xor\, b_i)\)

题解很神秘，看不懂。。。

两颗字典树dfs相互跑就好了

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

const int N = 1e5 + 10;

struct Trie {
	int root, sz, cnt[N * 31], nxt[N * 31][2];
	int newnode() {
		cnt[sz] = 0;
		memset(nxt[sz], 0, sizeof(nxt[sz]));
		return sz++; 
	}
	void init() {
		sz = 0;
		root = newnode();
	} 
	void insert(int x) {
		int p = root;
		for(int i = 29; ~i; --i) {
			int c = (x >> i) & 1;
			if(!nxt[p][c])
			nxt[p][c] = newnode();
			p = nxt[p][c];
			cnt[p]++;
		}
	}
}A, B;

int a[N], b[N], T, n;

vector< pair<int, int> > ans;

void dfs(int p1, int p2, int x, int s) {
	int c = min(A.cnt[p1], B.cnt[p2]);
	A.cnt[p1] -= c; B.cnt[p2] -= c;
	if(s == 30) {
		ans.push_back(make_pair(x, c));
		//cout << x << " " << c << endl;
		return;
	}
	if(A.cnt[A.nxt[p1][0]] && B.cnt[B.nxt[p2][0]])
		dfs(A.nxt[p1][0], B.nxt[p2][0], x << 1, s + 1);
	if(A.cnt[A.nxt[p1][1]] && B.cnt[B.nxt[p2][1]])
		dfs(A.nxt[p1][1], B.nxt[p2][1], x << 1, s + 1);
	if(A.cnt[A.nxt[p1][0]] && B.cnt[B.nxt[p2][1]])
		dfs(A.nxt[p1][0], B.nxt[p2][1], x << 1 | 1, s + 1);
	if(A.cnt[A.nxt[p1][1]] && B.cnt[B.nxt[p2][0]])
		dfs(A.nxt[p1][1], B.nxt[p2][0], x << 1 | 1, s + 1);
}

int main() {
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &n);
		A.init(); B.init();
		for(int i = 1; i <= n; ++i) {
			scanf("%d", &a[i]);
			A.insert(a[i]);
		}
		for(int i = 1; i <= n; ++i) {
			scanf("%d", &b[i]);		
			B.insert(b[i]);
		}
		ans.clear();
		dfs(A.root, B.root, 0, 0);
		sort(ans.begin(), ans.end());
		for(int i = 0; i < ans.size(); ++i) {
			for(int j = 1; j <= ans[i].second; ++j) {
				printf("%d", ans[i].first);
				if(j != ans[i].second)
					printf(" ");
			}
			if(i + 1 != ans.size())
				printf(" ");
		}
		puts("");
	}
	return 0;	
}

1004. equation

solved at 03:02(+2)

队友做的，没看

1005. permutation1

solved at 02:25(+1)

T组数据，你有一个\(1-n\)的排列\(a\)，定义\(d[i]=a[i+1]-a[i](1<=i<n)\)

求字典序第\(k\)小的\(d\)序列对应的原序列\(a\)

\(2<=n<=20, 1<=k<=min(factorial(n), 1e4), 1<=T<=40\)

题解是个搜索。。。

我的（实际上是队友的想法我的实现）做法是枚举\(d\)数组计算出剩下的方案数，复杂度\(O(Tn^3)\)

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

int a[25], T, n, k, b[25], vis[25][50], N = 25, s, ss, mx, mn;
long long f[25], tmp;

void get_a() {
	int x = 0, s = 0, p = 1;
	for(int i = 1; i < n; ++i) {
		s += b[i];
		if(s > x) {
			p = 1 + i;
			x = s;
		}
	}
	a[p] = n;
	for(int i = p + 1; i <= n; ++i) {
		a[i] = b[i - 1] + a[i - 1];
	}
	for(int i = p - 1; i > 0; --i) {
		a[i] = a[i + 1] - b[i];
	}
}

int main() {
	f[0] = f[1] = 1;
	for(int i = 2; i <= 20; ++i)
		f[i] = f[i - 1] * i;
	scanf("%d", &T);
	while(T--) {
		memset(vis, 0, sizeof(vis));
		scanf("%d%d", &n, &k);
		vis[0][N] = 1; s = 0; mx = 0; mn = 0;
		for(int i = 1; i < n; ++i) {
			for(int j = 1 - n; j <= n - 1; ++j) {
				if(vis[i - 1][N - j] || j == 0) continue;
				ss = s + j;
				bool flag = 0;
				for(int l = N - n; l <= N + n; ++l) {
					if(vis[i - 1][l] && (abs(l + j - N) >= n || l + j == N))
						flag = 1;
				}
				if(flag) continue;
				if(i == 1)
					tmp = f[n - i - 1] * (n - abs(ss));
				else 
					tmp = f[n - i - 1] * (n - max(max(abs(mx), abs(mn)), max(max(abs(ss - mx), abs(ss - mn)), max(abs(mx - mn), abs(ss)))));
				if(tmp >= k) {
					b[i] = j;
					s += j;
					int sp = 0;
					vis[i][sp + N] = 1;
					for(int l = i; l; --l) {
						sp += b[l];
						vis[i][sp + N] = 1;
					}
					mx = max(mx, ss);
					mn = min(mn, ss);	
					break;
				}
				else {
					k -= tmp;
				}
			}
		}
		get_a();
		for(int i = 1; i <= n; ++i)
			printf("%d%c", a[i], " \n"[i == n]);
	}
	return 0;
}

1006. string matching

solved at 00:28(+1)

没开long long wa了一发。。

exkmp裸题

1007. permutation2

solved at 00:57(+1)

有一个1-n的排列，你知道第一项和最后一项，且相邻项的差不超过2,求方案数

oeis+队友想的+瞎搞搞

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

const int N = 1e5 + 10, mod = 998244353;

int f[N], n, x, y, T;

int get(int x, int y) {
	if(x == y - 1) {
		if(x != 1 && y != n)
			return 0;
		return 1;
	}
	if(x != 1) x++;
	if(y != n) y--;
	return f[y - x];
}

int main() {
	f[0] = f[1] = f[2] = 1;
	for(int i = 3; i < N; ++i)
		f[i] = (f[i - 1] + f[i - 3]) % mod;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d%d", &n, &x, &y);
		printf("%d\n", get(x, y));
	}
	return 0;
}