武工大2022蓝桥杯预选赛题解复现

#include<bits/stdc++.h>     //万能头文件，费时但好用
using namespace std;

int main() {
	ios::sync_with_stdio(false);   //不向C语言兼容，加速用
	cin.tie(0);                    //io不绑定，加速用
	map<string, int> num;
	num["one"] = 1; num["two"] = 2; num["three"] = 3; num["four"] = 4; num["five"] = 5;
	num["six"] = 6; num["seven"] = 7; num["eight"] = 8; num["nine"] = 9; num["zero"] = 0;
	
	string s;
	vector<string> vs;

	while (cin >> s) {
		string s2 = "";
		for (auto ch : s) if (islower(ch)) s2 += ch;
		vs.push_back(s2);
	}

	int x = 1, nmb = 0, sum = 0;
	for (auto &s0 : vs) {
		cout << s0 << " ";
		if (s0 != "add" && s0 != "sub") nmb = nmb * 10 + num[s0];
		else {
			sum += nmb * x;
			nmb = 0;
			if (s0 == "add") x = 1;
			else x = -1;
		}
	}
	if (nmb != 0) sum += nmb * x;

	cout << "\n" << sum;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N = 110;
typedef pair<int, int> pii;

int n, m, k1, k2;
bool grid[N][N];                   //标记方格
int dx[] = { 0, 1, 0, -1 };        //控制上下左右移动
int dy[] = { 1, 0, -1, 0 };

int get(int x, int y) {            //获取数位之和
	int sum = 0;

	while (x) {
		sum += x % 10;
		x /= 10;
	}
	while (y) {
		sum += y % 10;
		y /= 10;
	}

	return sum;
}

int bfs() {
	grid[1][1] = true;
	queue<pii> q;
	q.push(make_pair(1, 1));
	int count = 1;

	while (!q.empty()) {
		pii p = q.front();       //每次循环以队首元素为中心
		q.pop();                 //及时将队首弹出

		for (int i = 0; i < 4; i++) {
			int x = p.first + dx[i];
			int y = p.second + dy[i];
			if (x <= n && x >= 1 && y <= m && y >= 1 && !grid[x][y]) {
				grid[x][y] = true;
				int num = get(x, y);
				if (num <= k2 && num >= k1) {
					q.push(make_pair(x, y));
                    //只有前辈元素在需求范围内才有机会加入队列
					count++;
				}
			}
		}
	}

	return count;
}

int main() {
	cin >> n >> m >> k1 >> k2;
	cout << bfs();
	return 0;
}

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

int x;          //全局变量，不用向函数传递参数
int a[4];

bool f() {
	for (int i = 0; i < 16; i++) {
		int sum = 0;
		for (int j = 0; j < 4; j++) {
			if (i >> j & 1) sum += a[j];
			else sum -= a[j];
		}
		if (sum == x) return true;
	}
	return false;
}

int main() {
	int n;
	cin >> n;

	while (n--) {
		for (int i = 0; i < 4; i++) cin >> a[i];
		cin >> x;
		if (f()) cout << "YES" << endl;     //若用 puts() 会方便一点
		else cout << "NO" << endl;
	}

	return 0;
}

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

int x;
int a[4];

bool f(int n, int m) {
	if (n == 4) return m == x;
	return f(n + 1, m + a[n]) || f(n + 1, m - a[n]);
}

int main() {
	int n;
	cin >> n;

	while (n--) {
		for (int i = 0; i < 4; i++) cin >> a[i];
		cin >> x;
		if (f(0, 0)) cout << "YES" << endl;
		else cout << "NO" << endl;
	}

	return 0;
}

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

int m, n;  //n是圆筒层数，m是盘子总数
long long r[300010], k[300010];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
    //实测这两句能让效率提升不少

	cin >> n >> m;

	for (int i = n; i > 0; i--) cin >> r[i];
    //便于从底端开始
	for (int i = 1; i <= m; i++) cin >> k[i];

	if (n < m) {
		cout << 0;
		return 0;
	}

	for (int i = 1; i <= m; i++)
		if (k[i] > r[n]) {
			cout << 0;
			return 0;
		}

	for (int i = n; i > 1; i--) {
		if (r[i] < r[i - 1]) r[i - 1] = r[i];
        //圆筒整形，统一向蛇精脸发展
	}

	int x = 1;
	for (int i = 1; i <= m; i++) {
		for (int j = x; j <= n; j++) {
			if (r[j] >= k[i]) {
				x = j + 1;
				break;
			}
		}
		if (x > n) {
			cout << 0;
			return 0;
		}
	}

	cout << n - x + 2;
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N = 2000010;  
//暂时没搞懂为什么要这么大，按题目应该1000000就够了。
using  ll = long long;
ll a[N], b[N];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n, m, k;

	cin >> n;
	for (int i = 0; i <= n; i++) cin >> a[i];
	cin >> m;
	for (int i = 0; i <= m; i++) cin >> b[i];
	cin >> k;

	ll sum = 0;
	for (int i = 0; i <= min(n, k); i++)
		sum = (sum + (a[i] % 998244353 * b[k - i] % 998244353) % 998244353) % 998244353;
	cout << sum;
	return 0;
}

int find(int x){return f[x]==x?x:(f[x]=find(f[x]));}

void join(int x,int y){father[find(x)]=find(y);}

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

const int INF = 0x3f3f3f3f;

int n, m, h;
int baol[100010];
struct load {
	int u, v, w;
} loads[200010];

bool f(load a, load b) { return a.w < b.w; };  //定义排序方法

int find(int x) { return baol[x] == x ? x : baol[x] = find(baol[x]); }
int kruskal() {
	int sum = 0, count = 1;
	for (int i = 0; i < n; i++) baol[i] = i;
	for (int i = 0; i < m; i++) {
		int a = loads[i].u, b = loads[i].v, c = loads[i].w;
		a = find(a);
		b = find(b);
		if (a != b) {
			baol[a] = b;
			sum += c;
			count++;
		}
	}

	if (count < n) return INF;
	return sum;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	cin >> n >> m >> h;
	for (int i = 0; i < m; i++) {
		cin >> loads[i].u >> loads[i].v >> loads[i].w;
	}

	sort(loads, loads + m, f);         //可使用 Lambda 表达式，如下
	//sort(loads, loads + m, [&](load a, load b) { return a.w < b.w; });
	if (kruskal() <= h) cout << "YES";
	else cout << "NO";

	return 0;
}