【Codeforces】Codeforces Round 872 (Div. 2) (ABC) (-38!)

原创已于 2023-05-14 15:56:39 修改 · 431 阅读

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于 2023-05-09 10:56:02 首次发布

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文章包含三道编程题目，分别涉及字符串的特性判断、矩阵中最大差值的计算以及特定场景下的布局优化问题。解题策略包括暴力搜索、排序和使用哈希表记录状态，同时利用前缀和优化计算过程。

A LuoTianyi and the Palindrome String

解题思路

暴力打表找规律，发现当字符串中只由1种构成就为-1，其他为字符串长度-1

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
ll T;
string s, y;
int ans;
ll len;
unordered_map<char, bool>mp;
void solve() {
	ans = -1;
	mp.clear();
	cin >> s;
	for (int i = 0; i < s.size(); i++) mp[s[i]] = 1;
	if (mp.size() == 1) ans = -1;
	else ans = s.size() - 1;
	cout << ans << endl;
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	T = 1;
	cin >> T;
	while (T--) solve();
	return 0;
}

B LuoTianyi and the Table

解题思路

第一点可以确定左上角要么是最大，要么是最小，为了使差值最大，那么次小或者次大就应该放在左上角的相邻格，两个相邻格中还得判断n和m的大小

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
const int N = 2e5 + 5;
ll T;
ll n, m;
ll ans1, ans2;
ll b[N];
void solve() {
	cin >> n >> m;
	for (int i = 1; i <= n * m; i++) cin >> b[i];
	sort(b + 1, b + n * m + 1);
	ll ma = b[n * m], mi1 = b[1], mi2 = b[2];
	ans1 = 0;
	ans1 += (ma - mi1) * (n - 1) * (m - 1);
	ans1 += (ma - max(mi1, mi2)) * (min(n, m) - 1);
	ans1 += (ma - min(mi1, mi2)) * (max(n, m) - 1);
	ll mi = b[1], ma1 = b[n * m], ma2 = b[m * n - 1];
	ans2 = 0;
	ans2 += (ma1 - mi) * (n - 1) * (m - 1);
	ans2 += (max(ma1, ma2) - mi) * (max(n, m) - 1);
	ans2 += (min(ma1, ma2) - mi) * (min(n, m) - 1);
	cout << max(ans1, ans2) << endl;
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	T = 1;
	cin >> T;
	while (T--) solve();
	return 0;
}

C LuoTianyi and the Show

解题思路

1.只用-1

2.只用-2

3.从中间考虑

对于第三种情况，用哈希把 $x>0$ 的情况存起来，用前后缀数组表示左和右的空位，再判断即可

当然也可以不用求前后缀，但是得排序

参考代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
const int N = 2e5 + 5;
ll T;
ll n, m;
ll x;
ll ans;
ll res;
unordered_map<ll, int>mp;
struct node {
	ll x;
	ll l, r;
};
void solve() {
	ans = 0;
	mp.clear();
	ll cl = 0, cr = 0;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> x;
		if (x == -1) cl++;
		else if (x == -2) cr++;
		if (x > 0) mp[x]++;
	}
	x = mp.size();
	ll fl[m + 5] = {};
	ll fr[m + 5] = {};
	for (int i = 1; i <= m; i++) {
		fl[i] = fl[i - 1];
		if (!mp[i]) fl[i]++;
	}
	for (int i = m; i >= 1; i--) {
		fr[i] = fr[i + 1];
		if (!mp[i]) fr[i]++;
	}

	for (ll i = 1; i <= m; i++) {
		if (mp[i]) {
			res = x;
			res += min(fl[i], cl);
			res += min(fr[i], cr);
			ans = max(ans, res);
		}
	}

	res = 0;
	for (int i = 1; i <= m; i++) {
		if (mp[i]) res++;
		else {
			if (cr > 0) {
				cr--;
				res++;
			}
		}
	}
	ans = max(ans, res);

	res = 0;
	for (int i = m; i >= 1; i--) {
		if (mp[i]) res++;
		else {
			if (cl > 0) {
				cl--;
				res++;
			}
		}
	}
	ans = max(ans, res);

	cout << ans << endl;
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	T = 1;
	cin >> T;
	while (T--) solve();
	return 0;
}

用数组把位置存起来

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
const int N = 2e5 + 5;
ll T;
ll n, m;
ll x;
ll ans;
ll res;
unordered_map<ll, int>mp;
vector<ll>v;
bool cmp(ll a, ll b) {
	return a < b;
}
void solve() {
	ans = 0;
	mp.clear();
	v.clear();
	ll cl = 0, cr = 0;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> x;
		if (x == -1) cl++;
		else if (x == -2) cr++;
		else mp[x]++;
	}
	x = mp.size();
	v.push_back(-1);
	for (auto i : mp) v.push_back(i.first);

	sort(v.begin(), v.end(), cmp);
	for (ll i = 1; i <= x; i++) {
		res = x + min(cl, v[i] - i) + min(cr, m - v[i]  - (x - i));
		ans = max(ans, res);
	}

	res = max(min(m, cl + x), min(m, cr + x));
	ans = max(ans, res);

	cout << ans << endl;
}
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	T = 1;
	cin >> T;
	while (T--) solve();
	return 0;
}