2021春季甲级PAT

本文最后更新于：星期四, 二月 3日 2022, 9:15 晚上

A

题意

给定一个n,MAXPn, MAXP，问在1−MAXP1-MAXP范围内是否存在一个长度为nn的并且全部由素数组成的等差序列，优先找差值最大的，若有多个则找最大值最大的，若不存在则输出范围内最大素数

解法

素数筛出范围内所有质数，暴力判断每个差值是否存在即可

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#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;

int n, m, k;
bool isPrime[N];
int prime[N];

void gao() {
	memset(isPrime, true, sizeof isPrime);
	for (int i = 2; i <= m; ++i) {
		if (isPrime[i]) {
			prime[++k] = i;
		}
		for (int j = 2 * i; j <= m; j += i) {
			isPrime[j] = false;
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	
	cin >> n >> m;
	gao();
	for (int dif = 100000; dif >= 1; --dif) {
		for (int i = k; i >= n; --i) {
			if (prime[i] - (n - 1) * dif <= 0) break;
			vector<int> v;
			v.push_back(prime[i]);
			int last = prime[i];
			while (true) {
				if (v.size() == n) {
					break;
				}
				if (isPrime[last - dif]) {
					last = last - dif;
					v.push_back(last);
				} else {
					break;
				}
			}
			if (v.size() == n) {
				reverse(v.begin(), v.end());
				for (int o = 0, len = v.size(); o < len; ++o) {
					cout << v[o] << " \n"[o == len - 1];
				}
				return 0;
			}
		}
	}
	cout << prime[k] << endl;
	return 0;
}

B

题意

有nn个人，每个人有进出lablab的申请时间，同一时间内只能有一个人在lablab中，问最多同意几个申请

解法

典型区间贪心，然而忘了怎么写

于是dpdp冲过去了

dpdp做法：

将开始时间排序，然后简单dpdp…只有当当前开始时间比上一次结束时间迟的时候才可以转移

区间贪心做法：

待补

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#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;

struct E {
	int begin, end;
	
	E() {}
	
	E(int begin, int end): begin(begin), end(end) {}

	bool operator < (const E &other) const {
		if (begin != other.begin) {
			return begin < other.begin;
		} else {
			return end < other.end;
		}
	}
}a[N];

int n;
int f[N];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		int h1, h2, m1, m2, s1, s2;
		scanf("%d:%d:%d %d:%d:%d", &h1, &m1, &s1, &h2, &m2, &s2);
		int t1 = h1 * 60 * 60 + m1 * 60 + s1;
		int t2 = h2 * 60 * 60 + m2 * 60 + s2;
		a[i] = E(t1, t2);
	}
	sort(a + 1, a + 1 + n);
	int res = 0;
	for (int i = 1; i <= n; ++i) {
		f[i] = max(1, f[i]);
		for (int j = i + 1; j <= n; ++j) {
			if (a[j].begin >= a[i].end) {
				f[j] = max(f[j], f[i] + 1);
			}
		}
		res = max(res, f[i]);
	}	
	printf("%d\n", res);
	return 0;
}

C

题意

要求维护一个heapheap，然后一堆无脑询问

思路

按照题意模拟即可

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#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;

int n, q;
int a[N];

void build(int x) {
	for (int i = x; i > 1; i /= 2) {
		int fa = i / 2;
		if (a[i] > a[fa]) {
			swap(a[i], a[fa]);
		} else {
			break;
		}
	}
}

int find(int x) {
	for (int i = 1; i <= n; ++i) {
		if (a[i] == x) return i;
	}
	return -1;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	
	cin >> n >> q;
	for (int i = 1; i <= n; ++i) {
		cin >> a[i];
		build(i);
	}
//	for (int i = 1; i <= n; ++i) {
//		cout << a[i] << endl;
//	}
	int x, y;
	string s;
	for (int cas = 1; cas <= q; ++cas) {
		cin >> x >> s;
		if (s == "and") {
			cin >> y;
			cin >> s >> s;
			x = find(x), y = find(y);
			if (x == -1 || y == -1) {
				cout << 0;
				continue;
			}
			if (x / 2 == y / 2) {
				cout << 1;
			} else {
				cout << 0;
			}
		} else {
			cin >> s >> s;
			if (s == "root") {
				x = find(x);
				if (x == 1) cout << 1;
				else cout << 0;
			} else if (s == "parent") {
				cin >> s >> y;
				x = find(x), y = find(y);
				if (x == -1 || y == -1) {
					cout << 0;
					continue;
				}
				if (x == y / 2) {
					cout << 1;
				} else {
					cout << 0;
				}
			} else if (s == "left") {
				cin >> s >> s >> y;
				x = find(x), y = find(y);
				if (x == -1 || y == -1) {
					cout << 0;
					continue;
				}
				if (y * 2 == x) {
					cout << 1;
				} else {
					cout << 0;
				}
			} else {
				cin >> s >> s >> y;
				x = find(x), y = find(y);
				if (x == -1 || y == -1) {
					cout << 0;
					continue;
				}
				if (y * 2 + 1 == x) {
					cout << 1;
				} else {
					cout << 0;
				}
			}
		}
	}
	cout << endl;
	return 0;
}

D

题意

给出一张n+1n+1个点的无向图，从00出发，每次找最近的点，如果有多个则找下标最小的，询问访问顺序和路径长度，如果无法走完所有的点则输出无法访问的点

思路

FloydFloyd之后dfsdfs即可

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#include <bits/stdc++.h>

using namespace std;

const int N = 2e2 + 10;
const int INF = 0x3f3f3f3f;

struct Edge {
	int v, w;
	
	Edge() {}
	
	Edge(int v, int w): v(v), w(w) {}
	
	bool operator < (const Edge &other) const {
		if (w != other.w) return w < other.w;
		else return v < other.v;
	}
};

int n, m;
vector<int> v;
bool vis[N];
int res;
int mp[N][N];
vector<Edge> G[N];

void Init() {
	memset(vis, 0, sizeof vis);
	memset(mp, 0x3f, sizeof mp);
	for (int i = 0; i <= n; ++i) mp[i][i] = 0;
}

void Floyd() {
	for (int k = 0; k <= n; ++k) {
		for (int i = 0; i <= n; ++i) {
			for (int j = 0; j <= n; ++j) {
				mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
			}
		}
	}
	for (int i = 0; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) if (i != j && mp[i][j] != INF) {
			G[i].push_back(Edge(j, mp[i][j]));
		}
		sort(G[i].begin(), G[i].end());
	}
}

void gao(int u) {
	vis[u] = true;
	v.push_back(u);
	for (auto it : G[u]) {
		int v = it.v, w = it.w;
		if (vis[v]) continue;
		res += w;
		gao(v);
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	
	cin >> n >> m;
	Init();
	for (int i = 1, u, v, w; i <= m; ++i) {
		cin >> u >> v >> w;
		mp[u][v] = w;
		mp[v][u] = w;
	}
	Floyd();
	gao(0);
	for (int i = 0, len = v.size(); i < len; ++i) {
		cout << v[i] << " \n"[i == len - 1];
	}
	if (v.size() == n + 1) {
		cout << res << endl;
	} else {
		vector<int> vec;
		for (int i = 0; i <= n; ++i) {
			if (!vis[i]) {
				vec.push_back(i);
			}
		}
		for (int i = 0, len = vec.size(); i < len; ++i) {
			cout << vec[i] << " \n"[i == len - 1];
		}
	}
	return 0;
}

PAT

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