Codeforces Global Round 10

A

#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(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 2e5 + 5;
 
int n, m, _, k;
ll a[N];
 
int main() {
    IO;
    for (cin >> _; _; --_) {
        cin >> n >> a[1];
		bool f = 1;
		rep (i, 2, n) {
			cin >> a[i];
			if (a[i] != a[i - 1]) f = 0;
		}
 
		if (n == 1) { cout << 1 << '\n'; continue; }
		else if (n == 2) { cout << (a[1] == a[2] ? 2 : 1) << '\n'; continue; }
		else if (!f) cout << 1 << '\n';
		else cout << n << '\n';
    }
    return 0;
}

B

#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(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 2e5 + 5;
 
int n, m, _, k;
ll a[N], c;
 
int main() {
    IO;
    for (cin >> _; _; --_) {
        cin >> n >> c >> a[1];
		int mx = 1, mi = 1;
		rep (i, 2, n) {
			cin >> a[i];
			if (a[i] > a[mx]) mx = i;
			if (a[i] < a[mi]) mi = i;
		}
 
		if (c & 1) {
			rep (i, 1, n) cout << a[mx] - a[i] << ' ';
			cout << '\n';
		} else {
			rep (i, 1, n) cout << a[i] - a[mi] << ' ';
			cout << '\n';
		} 
    }
    return 0;
}

C

记录上次一增加了多少, 可以无缝衔接给下次, 贪心

#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(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 2e5 + 5;
 
int n, m, _, k;
ll a[N];
 
int main() {
    IO;
    for (cin >> _; _; --_) {
        cin >> n >> a[1];
		ll mx = a[1], m = 0, ans = 0;
		rep (i, 2, n) {
			cin >> a[i];
			if (a[i] > mx) mx = a[i], m = 0;
			else if (a[i] == mx) m = 0;
			else {
				ll cur = mx - a[i] - m;
				if (cur > 0) ans += cur, m += cur;
				else m = mx - a[i];
			}
		}
		cout << ans << '\n';
    }
    return 0;
}

D

发现连续两个都是合法的, 所以对于 三个连续的 必须改一次

肯定是改连续的第三个字母啦

1.对于全是一种字母的, 每3次改一次, 对于最后多出来的 1 或 2个再改一次就好
2.混合的, 我们断环, 当然是从 s[i] != s[i - 1] 处断环, 使得字母每段都是连续的
还是每3次改一次, 但是对于每段多出来的 1 或 2个, 比如 RRR RR L 发现由于余出的 RR 左边变成了 L, 并不用修改 (相当于 RRR -> RRL, 已经被改好了)
即, 每段长度 / 3即可

#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(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 2e5 + 5;
 
int n, m, _, k;
string s;
int main() {
    IO;
    for (cin >> _; _; --_) {
        cin >> n >> s;
		int f = -1;
		rep (i, 1, n - 1) {
			s += s[i - 1];
			if (s[i] != s[i - 1]) {
				f = i;
				break;
			}
		}
 
		if (f == -1) cout << n / 3 + (n % 3 != 0) << '\n';
		else {	
			int ans = 0, cnt = 1;
			rep (i, f + 1, f + n - 1) 
				if (s[i] == s[i - 1]) ++cnt;
				else ans += cnt / 3, cnt = 1;
 
			cout << ans + cnt / 3 << '\n';
		}
    }
    return 0;
}

E

每个位置无非是从 上和左走来, 所以确保 次对角线 上的每个点的 数值范围 不相交 即可

#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(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 25;
 
int n, m, _, k;
ll a[N][N];
PII ans[N << 1];
PLL b[N][N];
 
int main() {
	IO; cin >> n;
	rep (k, 1, (n - 1) << 1) {
		ll cur = -1;
		rep (i, max(0, k - n + 1), min(k, n - 1)) {
			int j = k - i;
			ll mn = 1e18, mx = -1e18;
			if (i) mn = min(mn, b[i - 1][j].fi), mx = max(mx, b[i - 1][j].se);
			if (j) mn = min(mn, b[i][j - 1].fi), mx = max(mx, b[i][j - 1].se);
 
			b[i][j] = {cur + 1, cur += 1 + mx - mn};
			a[i][j] = b[i][j].fi - mn;
		}
	}
	
	rep (i, 0, n - 1) {
		rep (j, 0, n - 1) cout << a[i][j] << ' ';
		cout << '\n';
	}
	
	cout.flush();
	cin >> m;
	rep (i, 1, m) {
		ll k; cin >> k;
		int x = n - 1, y = n - 1; 
		per (i, n - 1 << 1, 0) {
			ans[i] = {x + 1, y + 1};
			k -= a[x][y];
			if (x && k >= b[x - 1][y].fi && k <= b[x - 1][y].se) --x;
			else --y;
		}
		rep (i, 0, n - 1 << 1) cout << ans[i].fi << ' ' << ans[i].se << '\n';
		cout << '\n';
		cout.flush();
	}
	return 0;
}

F

给的序列递增, 只有在 \(a_1, a_1 + 1, ..., a_1 + n - 1\) 的情况下, 泥土不在移动

只要高层泥土能移动, 必然顺着下滑, 故, 泥土最终会使得, 原序列变为, \(a_1 + x, a_1 + x + 1, ..., a_1 + x + m - 1, a_1 + x + m - 1, a_{ m + 1}, .. a_n\)

#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(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;

const int N = 1e6 + 5;

int n, m, _, k;
ll a[N];

int main() {
    IO;
    cin >> n >> a[1]; ll sum = 0;
    rep (i, 2, n) cin >> a[i], sum += a[i] - i + 1 - a[1];
    ll x = sum / n, y = sum % n;
    rep (i, 1, n) cout << a[1] + x + (y > 0) + i - 1 << ' ', --y;
    cout << '\n';
    return 0;
}