Codeforces Round 481 (Div. 3) 补题AK

比赛链接

A 模拟

思路

直接采用哈希去重即可

代码

#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;

int n;
int a[N];
umap<int, int> mp;
vector<int> ans;
void solve() {
	cin >> n;
	forn(i, 1, n) cin >> a[i];

	forn_(i, n, 1) {
		if (mp[a[i]] == 0) ans.push_back(a[i]);
		mp[a[i]]++;
	}

	cout << ans.size() << endl;
	reverse(ans.begin(), ans.end());
	for (auto i : ans) cout << i << ' ';
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//cin >> T;
	T = 1;

	while (T--) solve();
	return 0;
}

B 模拟 贪心

思路

考虑贪心，为了删除次数最少，肯定只能删x，只要连续x长度小于3就行了

找到连续的x然后判断长度该删多少，累加答案即可

代码

#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;

int n;
char s[N];
int ans;
void solve() {
	cin >> n;
	forn(i, 1, n) cin >> s[i];

	int res = 0;
	s[0] = s[n + 1] = '?';
	forn(i, 0, n + 1) {
		if (s[i] != 'x') {
			ans += max(0, res - 2);
			res = 0;
		} else res++;
	}

	cout << ans;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//cin >> T;
	T = 1;

	while (T--) solve();
	return 0;
}

C 二分

思路

代码

#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
ll T;

ll n, m, a;
ll s[N];

ll find() {
	ll l = 0, r = n;
	while (l < r) {
		ll mid = l + r >> 1;
		if (s[mid] >= a) r = mid;
		else l = mid + 1;
	}
	return l;
}
void solve() {
	cin >> n >> m;
	forn(i, 1, n) cin >> a, s[i] = s[i - 1] + a;

	forn(i, 1, m) {
		cin >> a;
		ll f = find();
		cout << f << ' ' << a - s[f - 1] << endl;
	}
}

int main() {
	//cin >> T;
	T = 1;

	while (T--) {
		solve();
	}

	return 0;
}

D 思维 枚举

思路

代码

#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;

int n;
int a[N];
int ans = inf;
void solve() {
	cin >> n;
	forn(i, 1, n) cin >> a[i];
	if (n == 1 || n == 2) {
		cout << 0;
		return;
	}

	int x = a[1], y = a[2];
	forn(i, -1, 1) forn(j, -1, 1) {
		a[1] = x + i, a[2] = y + j;
		int d = a[2] - a[1];
		int ok = 1, res = 0, lst = a[1];
		if (i != 0) res++;
		if (j != 0) res++;
		forn(k, 2, n) {
			int da = a[k] - lst;
			if (da == d) lst = a[k];
			else if (da == d + 1) res++, lst = a[k] - 1;
			else if (da == d - 1) res++, lst = a[k] + 1;
			else ok = 0;
		}
		if (ok) ans = min(ans, res);
	}

	if (ans == inf) ans = -1;
	cout << ans;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//cin >> T;
	T = 1;

	while (T--) solve();
	return 0;
}

E 数学

思路

就是解不等式组，答案就是区间长度，注意特判负数为0 

代码

#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;

int n, w;
ll a[N];
ll l, r;
void solve() {
	cin >> n >> w;
	forn(i, 1, n) cin >> a[i];
	l = 0, r = w;

	ll res = 0;
	forn(i, 1, n) {
		res -= a[i];
		l = max(l, res);
		r = min(r, w + res);
	}

	cout << max(0ll, r - l + 1) << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//cin >> T;
	T = 1;

	while (T--) solve();
	return 0;
}

F 二分 排序

思路

代码

#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;

int n, m;
int a[N], b[N];
int ans[N];
void solve() {
	cin >> n >> m;
	forn(i, 1, n) cin >> a[i], b[i] = a[i];

	sort(b + 1, b + n + 1);
	forn(i, 1, n) {
		int l = lower_bound(b + 1, b + n + 1, a[i]) - b;
		ans[i] = l - 1;
	}

	while (m--) {
		int x, y;
		cin >> x >> y;
		if (a[x] < a[y]) ans[y]--;
		if (a[x] > a[y]) ans[x]--;
	}

	forn(i, 1, n) cout << ans[i] << ' ';
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//cin >> T;
	T = 1;

	while (T--) solve();
	return 0;
}

G 模拟 贪心

思路

水题啊

肯定从最早的考试开始准备，就先以d为关键字排个序

就考虑贪心，从s开始填到d-1，如果填的数量少于c说明不符合题意，输出-1

代码

#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int T;

int n, m;
struct node {
	int s, d, c;
	int id;
} a[N];
bool cmp(node x, node y) {
	return x.d < y.d;
}
int ans[N];
void solve() {
	cin >> n >> m;
	forn(i, 1, m) cin >> a[i].s >> a[i].d >> a[i].c, a[i].id = i, ans[a[i].d] = m + 1;

	sort(a + 1, a + m + 1, cmp);
	forn(i, 1, m) {
		int s = a[i].s, d = a[i].d, c = a[i].c, id = a[i].id;
		if (d - s < c) {
			cout << -1;
			return;
		}
		int cnt = 0;
		forn(j, s, d - 1) {
			if (ans[j] != 0) continue;
			cnt++;
			ans[j] = id;
			if (cnt == c) break;
		}
		if (cnt < c) {
			cout << -1;
			return;
		}
	}

	forn(i, 1, n) cout << ans[i] << ' ';
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//cin >> T;
	T = 1;

	while (T--) solve();
	return 0;
}