MX 炼石计划 NOIP 模拟赛20（我真做过1~19吗？）

MX 炼石计划 NOIP 模拟赛 #20

T1 邻间的骰子之舞

二分答案，发现性质，签到，过。

记得开 __int128

没开，挂 30.

码：

CODE

#include <bits/stdc++.h>
typedef long long ll; 
using namespace std; 
const int N = 2e5 + 100; 

#ifdef linux
#define getchar getchar_unlocked
#define putchar putchar_unlocked
#endif
inline int read() {
	int x = 0; char c = getchar(); 
	while (c < '0' || c > '9') c = getchar(); 
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); 
	return x; 
}
template <typename T> void write(T x) {
	if (x / 10) write(x / 10); 
	putchar(x % 10 + '0'); 
}

ll n, x, y; 
ll maxer; 

inline bool check(__int128 mid) {
	for (__int128 i = 0; i <= maxer; ++ i) {
		if (i * x >= mid) break; 
		if (i == 0) {
			if ((mid - i * x) / y + 1 > n) return true; 
			continue; 
		}

		__int128 j = (mid - i * x) / y, val = 1, aver = j / (i + 1), last = j - aver * (i + 1); 
		for (ll k = 0; k <= i; ++ k) {
			if (last > 0) {
				val += (aver + 1) * val, last --; 
			} else val += aver * val; 
			if (val > n) return true; 
		}
	}

	return false; 
}

signed main() {
	freopen("dice.in", "r", stdin); 
	freopen("dice.out", "w", stdout); 
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); 
	cin >> n >> x >> y; 

	__int128 l = 1, r = 1e19, ans = 0; maxer = log2(n); 

	while (l <= r) {
		__int128 mid = ((__int128)(l + r) >> 1ll); 

		if (check(mid)) {
			ans = mid, r = mid - 1; 
		} else l = mid + 1; 
	}

	write(ans + x); 
}

T2 星海浮沉录

签到2，拿 \(set\) 维护一下每种数位置以及两者之间距离，然后线段树上二分。

码：

CODE

#include <bits/stdc++.h>
typedef long long ll; 
using namespace std; 
const int N = 5e5 + 100; 

#ifdef linux
#define getchar getchar_unlocked
#define putchar putchar_unlocked
#endif
inline int read() {
	int x = 0; char c = getchar(); 
	while (c < '0' || c > '9') c = getchar(); 
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); 
	return x; 
}
template <typename T> void write(T x) {
	if (x / 10) write(x / 10); 
	putchar(x % 10 + '0'); 
}

int n, q; int a[N]; 
int last[N]; 
multiset <int> num[N], dis[N]; 

#define lson (id << 1)
#define rson (id << 1 | 1)
#define mid ((l + r) >> 1)

int t[N << 2]; 
void updata(int id, int l, int r, int x, int c) {
	if (l == r) return t[id] = c, void(); 
	return ((x <= mid) ? updata(lson, l, mid, x, c) : updata(rson, mid + 1, r, x, c)), t[id] = max(t[lson], t[rson]), void(); 
}
int Query(int id, int l, int r, int k) {
	if (l == r) return l; 
	if (t[lson] >= k) return Query(lson, l, mid, k); 
	else return Query(rson, mid + 1, r, k); 
}

#undef mid

#define iter set <int>::iterator

inline void change(int x, int i, int j) {
	iter it = num[x].lower_bound(i), it2 = ++ it; -- it; 

	if (it != num[x].begin()) {
		int fir = *(-- it); ++ it; 
		dis[x].erase(dis[x].lower_bound(i - fir - 1)), dis[x].insert(j - fir - 1); 
	}
	if (it2 != num[x].end()) {
		int sec = *it2; 
		dis[x].erase(dis[x].lower_bound(sec - i - 1)), dis[x].insert(sec - j - 1); 
	}

	num[x].erase(i), num[x].insert(j); 
	updata(1, 0, n + 1, x, *dis[x].rbegin()); 
}

signed main() {
	freopen("star.in", "r", stdin); 
	freopen("star.out", "w", stdout); 
	n = read(), q = read(); 
	for (int i = 0; i <= n; ++ i) num[i].insert(0); 
	for (int i = 1; i <= n; ++ i) {
		a[i] = read(), num[a[i]].insert(i); 
		dis[a[i]].insert(i - last[a[i]] - 1); 
		last[a[i]] = i; 
	}
	for (int i = 0; i <= n; ++ i) {
		dis[i].insert(n - *num[i].rbegin()); num[i].insert(n + 1); 
		updata(1, 0, n + 1, i, *dis[i].rbegin()); 
	}

	int opt, x, tot = 0; 
	while (q --) {
		opt = read(), x = read(); 

		if (opt == 1) {
			if (a[x] == a[x + 1]) continue; 
			change(a[x], x, x + 1); 
			change(a[x + 1], x + 1, x); 
			swap(a[x], a[x + 1]); 
		} else {
			write(Query(1, 0, n + 1, x)), puts(""); 
		}
	}
}

注：byd梦熊还有什么逆天文件问题，甚至还只有一个测试点有文件问题？唐。

T3 勾指起誓

不会，咕了。

T4 _八交响曲

赛时没想出来。

考虑两个有序数组合并，只需要将镜像位置交换之后，发现其左右儿子区间保证左大于右或右大于左这么一个情况，遍历他儿子的时候就可以按位置交换了，非常优秀。

CODE

#include <bits/stdc++.h>
typedef long long ll; 
using namespace std; 
const int N = 200; 

int n; int m; 
vector <pair <int, int> > v[N]; 
int pos[N][N], pos2[N]; 

void Solve2(int l, int r, int dep, int ac) {
	if (l == r) return; 
	if (!pos[ac][dep]) pos[ac][dep] = ++ m; 
	int mid = (l + r) >> 1; 
	for (int i = l; i <= mid; ++ i) {
		int now = mid + i - l + 1; 
		if (now <= r && now != i) v[pos[ac][dep]].push_back(make_pair(i, now)); 
	}

	Solve2(l, mid, dep + 1, ac), Solve2(mid + 1, r, dep + 1, ac); 
}

void Solve(int l, int r, int dep) {
	if (l == r) return; 
	int mid = (l + r) >> 1; 
	Solve(l, mid, dep + 1), Solve(mid + 1, r, dep + 1); 
	if (!pos2[dep]) pos2[dep] = ++ m; 
	for (int i = 1; i <= mid; ++ i) 
		if (l + i - 1 < r - i + 1)
			v[pos2[dep]].push_back(make_pair(l + i - 1, r - i + 1)); 
	Solve2(l, mid, dep + 1, dep), Solve2(mid + 1, r, dep + 1, dep); 
}

signed main() {
	freopen("symphony.in", "r", stdin); 
	freopen("symphony.out", "w", stdout); 
	cin >> n; int p = n; 
	n = pow(2, ceil(log2(n))); 
	Solve(1, n, 1); 
	cout << m << '\n'; 

	for (int i = 1; i <= m; ++ i) {
		for (auto j : v[i]) {
			if (j.first <= p && j.second <= p)
			printf("CMPSWP R%d R%d ", j.first, j.second); 
		} puts(""); 
	}
}

总结

70(-30) + 100(-0) + 12(-0) + 10(-0)

中学生誓词

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