Codeforces Round #662 (Div. 2)

A

题看不懂,猜答案

多试试, A题嘛, 总能猜出来(没了100分), 实在是看不懂题

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
typedef unsigned long long ull;
typedef unsigned int uint;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[N], b[N];
 
int main() {
	IO;
	for (cin >> _; _; --_) {
		cin >> n;
		cout << n / 2 + 1 << '
';
	}
	return 0;
}

B

wa 1发, 细节没处理好

每次是弹操作, 我们直接根据改变之后长为 i 板子的数量, 维护两个值就好了

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
typedef unsigned long long ull;
typedef unsigned int uint;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[N], x, y;
char s[5];
 
int main() {
    IO;
	cin >> n;
	rep (i, 1, n) cin >> m, ++a[m];
	rep (i, 1, 1e5) {
		x += a[i] / 4;
		y += a[i] % 4 / 2;
	}
 
	cin >> n;
	rep (i, 1, n) {
		cin >> s >> m;
		if (s[0] == '+') {
			++a[m];
			if(a[m] % 4 == 0) ++x, --y;
			else if (a[m] % 4 == 2) ++y;
		}
		else {
			--a[m];
			if (a[m] % 4 == 3) --x, ++y;
			else if (a[m] % 4 == 1) --y;
		}
 
		if (x && (x > 1 || y >= 2)) cout << "YES
";
		else cout << "NO
";
	}
    return 0;
}

C

首先考虑种类最多的馅饼, 占据 a 个位置, 那么之间的间隙最大值为 x = (n - a) / (a - 1),

因为是向下取整, 那么我们记下余数(一会用) y = n - a - (a - 1) * x

考虑其他的, 全是数量小于a的, 那么答案必是 a

如果存在 b 种馅饼数量和 a 相等
b <= y, 毫无疑问错位满足 x, 答案是 x
b > y, 明显 x 大了, --x, 循环重判(y就变大了)

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
typedef unsigned long long ull;
typedef unsigned int uint;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[N], b[N];
 
int main() {
	IO;
	for (cin >> _; _; --_) {
		cin >> n;
		memset(b, 0, sizeof b);
		rep(i, 1, n) cin >> a[i], ++b[a[i]];
		sort(b + 1, b + 1 + n, greater<int>());
		int x = (n - b[1]) / (b[1] - 1), y = n - b[1] - (b[1] - 1) * x;
		while (true) {
			int t = 2;
			while (t <= n && b[t] == b[t - 1] && y) --y, ++t;
			if (t > n || b[t] != b[t - 1]) { cout << x << '
'; break; }
			y = n - b[1] - (b[1] - 1) * (--x);
		}
	}
	return 0;
}

D

拼接就完事了, l是left, r是right, u是up, d是down

add: 看了题解, 真的服气, 直接dp一次即可, 我这浪费时间了,tcl

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
typedef unsigned long long ull;
typedef unsigned int uint;
 
const int N = 2005;
 
int n, m, _, k;
char s[N][N];
int l[N][N], r[N][N], u[N][N], d[N][N];
int lu[N][N], ru[N][N], ld[N][N], rd[N][N];
 
int main() {
	IO;
	cin >> n >> m;
	ll ans = 0;
	rep(i, 1, n) {
		cin >> s[i] + 1;
 
		rep(j, 1, m) {
			if (s[i][j] == s[i][j - 1]) l[i][j] = l[i][j - 1] + 1;
			else l[i][j] = 1;
 
			if (s[i][j] == s[i - 1][j]) u[i][j] = u[i - 1][j] + 1;
			else u[i][j] = 1;
 
			if (s[i][j] == s[i - 1][j - 1]) lu[i][j] = min(min(u[i][j], l[i][j]), lu[i - 1][j - 1] + 2);
			else lu[i][j] = max(1, min(2, min(u[i][j], l[i][j])));
		}
	}
 
	per(i, n, 1) {
		per(j, m, 1) {
			if (s[i][j] == s[i][j + 1]) r[i][j] = r[i][j + 1] + 1;
			else r[i][j] = 1;
 
			if (s[i][j] == s[i + 1][j]) d[i][j] = d[i + 1][j] + 1;
			else d[i][j] = 1;
 
			if (s[i][j] == s[i + 1][j + 1]) rd[i][j] = min(min(r[i][j], d[i][j]), rd[i + 1][j + 1] + 2);
			else rd[i][j] = max(1, min(2, min(r[i][j], d[i][j])));
		}
 
		rep(j, 1, m) {
			if (s[i][j] == s[i + 1][j - 1]) ld[i][j] = min(min(l[i][j], d[i][j]), ld[i + 1][j - 1] + 2);
			else ld[i][j] = max(1, min(2, min(l[i][j], d[i][j])));
		}
	}
 
	rep(i, 1, n) {
		per(j, m, 1) {
			if (s[i][j] == s[i - 1][j + 1]) ru[i][j] = min(min(r[i][j], u[i][j]), ru[i - 1][j + 1] + 2);
			else ru[i][j] = max(1, min(2, min(r[i][j], u[i][j])));
 
			int c = min(min(lu[i][j], ru[i][j]), min(ld[i][j], rd[i][j]));
			ans += c;
		}
	}
 
	cout << ans;
	return 0;
}

E1

时间不够了, 没调出来, wa在test6

暴力思路, map记忆化搜索, 在截取串的时候, 用两个for, 把删去的第i个字母隔开就行了 O(len(s))

明天试试能不能暴力出来吧