【2019第四届团体程序设计天梯赛】个人题解（C++）

最新推荐文章于 2025-04-23 23:44:20 发布

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文章提供了一系列编程题目的解题思路和参考代码，包括模拟、算术运算等方法，主要针对初级和中级难度的算法问题，如字符串处理、数学计算和逻辑判断等。

L1-1 PTA使我精神焕发

解题思路：无

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e2 + 5;

int main() {
	cout << "PTA shi3 wo3 jing1 shen2 huan4 fa1 !";
	return 0;
}

L1-2 6翻了

解题思路：模拟即可

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e2 + 5;
vector<string>v;
string s;
int main() {
	getline(cin, s);
	string x;
	int l = 0;
	for (int i = 0; i < s.size(); i++) {
		if (s[i] == '6') {
			l++;
		} else {
			if (l) {
				if (l > 9) {
					cout << "27";
				} else if (l > 3) {
					cout << "9";
				} else {
					for (int j = 1; j <= l; j++) cout << "6";
				}
				l = 0;
			}
			cout << s[i];
		}
	}
	if (l) {
		if (l > 9) {
			cout << "27";
		} else if (l > 3) {
			cout << "9";
		} else {
			for (int j = 1; j <= l; j++) cout << "6";
		}
		l = 0;
	}
	cout << endl;
	return 0;
}

L1-3 敲笨钟

解题思路：模拟

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e2 + 5;
int n;
string s;
int main() {
	cin >> n;
	getchar();
	while (n--) {
		getline(cin, s);
		if (s.find("ong,") != -1 && s.find("ong.") != -1) {
			int position = 0;
			int icount = 1;
			while ((position = s.find(" ", position)) != -1) {
				position++;
				icount++;
			}
			icount -= 3;
			for (int i = 0; i < s.length(); i++) {
				if (s[i] == ' ') {
					icount--;
				}
				if (icount) {
					cout << s[i];
				} else {
					cout << " qiao ben zhong." << endl;
					break;
				}
			}
		} else {
			cout << "Skipped" << endl;
		}
	}
	return 0;
}

L1-4 心理阴影面积

解题思路：小学算术

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e2 + 5;
int main() {
	int x, y;
	cin >> x >> y;
	cout << 100 * 100 / 2 - (100 - x)*(100 - y) / 2 - x*y / 2 - (100 - x)*y;
	return 0;
}

L1-5 新胖子公式

解题思路：小学算术

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e2 + 5;
int main() {
	double k, m;
	cin >> k >> m;
	double x = k / (m * m);
	printf("%.1lf\n", x);
	if (x > 25) cout << "PANG\n";
	else cout << "Hai Xing\n";
	return 0;
}

L1-6 幸运彩票

解题思路：模拟

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e2 + 5;
int main() {
	int n;
	cin >> n;
	string s;
	while (n > 1) {
		n--;
		cin >> s;
		s = ' ' + s;
		int sum1 = 0, sum2 = 0;
		for (int i = 1; i <= 3; i++) sum1 += s[i] - '0';
		for (int i = 4; i <= 6; i++) sum2 += s[i] - '0';
		if (sum1 == sum2) cout << "You are lucky!\n";
		else cout << "Wish you good luck.\n";
	}
	cin >> s;
	s = ' ' + s;
	int sum1 = 0, sum2 = 0;
	for (int i = 1; i <= 3; i++) sum1 += s[i] - '0';
	for (int i = 4; i <= 6; i++) sum2 += s[i] - '0';
	if (sum1 == sum2) cout << "You are lucky!";
	else cout << "Wish you good luck.";
	return 0;
}

L1-7

解题思路：

参考代码

L1-8

解题思路：

参考代码

L2-1 特立独行的幸福

解题思路：用图把各个数字的关系存起来，入度为的0的即为特立独行的幸福数

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e4 + 5;
int f(int x) {
	int sum = 0;
	while (x > 0) {
		sum += pow(x % 10, 2);
		x /= 10;
	}
	return sum;
}
bool g[N][N];
bool is[N];
bool check(int x) {
	int cnt = 0;
	while (1) {
		cnt++;
		if (x == 1) break;
		if (cnt > 10) return 1;
		g[x][f(x)] = 1;
		x = f(x);
	}
	return 0;
}
bool is_p[N];
vector<int>pri(1);
bool iss[N];
void init() {
	for (int i = 1; i <= 10000; i++) {
		if (check(i)) is[i] = 1;
	}
	for (int i = 2; i <= 10000; i++) {
		if (iss[i] == 0) pri.push_back(i);
		for (int j = 1; j < pri.size() && i * pri[j] <= 10000; j++) {
			iss[i * pri[j]] = 1;
			if (i % pri[j] == 0) break;
		}
	}
	for (int i = 1; i < pri.size(); i++) is_p[pri[i]] = 1;
}
int main() {
	init();
	int a, b;
	cin >> a >> b;
	int ok = 0;
	for (int i = a; i <= b; i++) {
		if (is[i] == 0) {
			ok = 1;
			int t = 1;
			for (int j = a; j <= b; j++) {
				if (g[j][i] == 1) {
					t = 0;
					break;
				}
			}
			if (t) {
				cout << i << ' ';
				int cnt = 0;
				int x = i;
				while (1) {
					cnt++;
					x = f(x);
					if (x == 1) break;
				}
				if (is_p[i]) cnt *= 2;
				cout << cnt << endl;
			}
		}
	}
	if (!ok) cout << "SAD";
	return 0;
}

L2-2

解题思路：

参考代码

L2-3 深入虎穴

解题思路：找到入口，从入口开始BFS找最长路即可

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 5;
int n, k, x;
vector<int>g[N];
int dis;
int ans;
struct node {
	int x;
	int t;
};
void bfs(int st) {
	int res = 0;
	queue<node>q;
	q.push({st, 1});
	while (!q.empty()) {
		node now = q.front();
		q.pop();
		if (res < now.t) {
			res = now.t;
			ans = now.x;
		}
		for (int i = 0; i < g[now.x].size(); i++) {
			q.push({g[now.x][i], now.t + 1});
		}
	}
}
bool is[N];
int main() {
	int st;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &k);
		for (int j = 1; j <= k; j++) {
			scanf("%d", &x);
			is[x] = 1;
			g[i].push_back(x);
		}
	}
	for (int i = 1; i <= n; i++) {
		if (is[i] == 0) {
			st = i;
			break;
		}
	}
	bfs(st);
	cout << ans;
	return 0;
}

L2-4 彩虹瓶

解题思路：使用栈模拟

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e3 + 5;
int n, m, k;
int a[N];
bool check() {
	int cnt = 1;
	stack<int>s;
	for (int i = 1; i <= n; i++) {
		if (a[i] == cnt) {
			cnt++;
		} else {
			while (!s.empty() && s.top() == cnt) {
				s.pop();
				cnt++;
			}
			if (s.empty()) s.push(a[i]);
			else {
				if (s.top() < a[i]) return 0;
				else {
					if (s.size() == m) return 0;
					s.push(a[i]);
				}
			}
		}
	}
	return 1;
}
int main() {
	cin >> n >> m >> k;
	while (k--) {
		for (int i = 1; i <= n; i++) cin >> a[i];
		if (check()) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
}