AtCoder Beginner Contest 328

A - Not Too Hard (atcoder.jp)

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	i64 n, x, ans = 0;
	cin >> n >> x;
	vector<int> a(n);

	for (auto &i : a) {
		cin >> i;
		if (i <= x) ans += i;
	}

	cout << ans << '\n';

	return 0;

}

B - 11/11 (atcoder.jp)

只有月份和日期都是由同一个数字组成时,才可能有重期日

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	int N;
	cin >> N;
	vector<int> D(N + 1);

	i64 ans = 0;
	for (int i = 1; i <= N; i ++) {
		cin >> D[i];
		int op = i % 10;
		if (i < 10) {
			while (op <= D[i]) {
				ans ++;
				op = op * 10 + op;
			}
		} else if (i < 100 && i % 10 == i / 10) {
			while (op <= D[i]) {
				ans ++;
				op = op * 10 + op;
			}
		}
	}

	cout << ans << '\n';

	return 0;

}

C - Consecutive (atcoder.jp)

前缀和处理一下相邻相同的个数即可

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	int N, Q;
	string s;
	cin >> N >> Q >> s;
	vector<int> pre(N + 1);
	s = " " + s;
	for (int i = 1; i <= N; i ++) {
		pre[i] = pre[i - 1];
		if (s[i] == s[i - 1])pre[i] ++;
	}

	while (Q--) {
		int l, r;
		cin >> l >> r ;
		cout << pre[r] - pre[l] << '\n';
	}

	return 0;

}

D - Take ABC (atcoder.jp)

貌似正解是用栈来做,不过我直接模拟也能过,就是时间卡的有点极限

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	string s;
	cin >> s;
	int pos = 0;
	while (pos < s.size()) {
		if (s.substr(pos, 3) == "ABC") {
			s.erase(pos, 3);
			pos --;
			pos = max(pos, 0);
			if (s[pos] == 'B') pos --;
			pos = max(pos, 0);
			continue;
		}
		pos ++;
	}

	cout << s << '\n';
	return 0;

}

E - Modulo MST (atcoder.jp)

不能直接用克鲁斯卡尔算法去做

因为克鲁斯卡尔是按照固定权值下的计算最小生成树,但是这题是要求取模意义下的最小生成树,按照权值排序得到的结果不一定是最小的,不过观察到这题的范围很小,所以可以直接暴力去做

考虑对\(M\)条边选出\(N-1\)条边出来,多余就\(return\),如果在合并的过程中发现是个环也\(return\),剩下的就是比较了

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

struct Edge {
	i64 v, u, w;
};

constexpr int N = 10;
struct UFS {
    int p[N], rank[N], sz; /
    void link(int x, int y) {
        if (x == y)
            return;
        if (rank[x] > rank[y]) 
            p[y] = x;
        else
            p[x] = y;
        if (rank[x] == rank[y]) 
            rank[y]++;
    }
    void init(int n) {  
        sz = n;
        for (int i = 0; i <= sz; i++) {
            p[i] = i;
            rank[i] = 0;
        }
    }
    int find(int x) {
        return x == p[x] ? x : (p[x] = find(p[x])); 
    }
    void unin(int x, int y) {
        link(find(x), find(y));
    }
    void compress() {
        for (int i = 0; i < sz; i++)
            find(i);
    }
};

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

	i64 N, M, K;
	cin >> N >> M >> K;
	vector<Edge> g(M);
	for (int i = 0; i < M; i ++) {
		cin >> g[i].u >> g[i].v >> g[i].w;
		g[i].w %= K;
	}

	i64 ans = K;
	vector<bool> vis(M); 

	auto dfs = [&](auto self,int edge,int step){		
		if(step >= M){
			if(edge != N - 1) return ;

			UFS ufs;
			ufs.init(N);

			i64 sum = 0;
			for(int i = 0;i < M;i ++){
				if(!vis[i]) continue;
				int u = ufs.find(g[i].u);
				int v = ufs.find(g[i].v);
				if(u == v) return ;
				ufs.link(u,v);
				sum = (sum + g[i].w) % K;
			}

			ans = min(ans, sum % K);
			return ;
		}

		vis[step] = true;
		self(self, edge + 1, step + 1);
		vis[step] = false;
		self(self, edge, step + 1);
	};

	dfs(dfs,0,0);

	cout << ans << '\n';

	return 0;
}