Codeforces Round #589 (Div. 2)

Contest Info
Solutions

Contest Info

[Practice Link](https://codeforces.com/contest/1228)

Solved	A	B	C	D	E	F
5/6	O	O	O	O	Ø	-

O 在比赛中通过
Ø 赛后通过
! 尝试了但是失败了
- 没有尝试

Solutions

A. Distinct Digits

签到。

B. Filling the Grid

签到。

C. Primes and Multiplication

题意：

定义\(prime(x)\)为\(x\)的所有质因子构成的集合。

定义\(g(x, p)\)为最大的\(p^k\)满足\(p^k \;|\; x\)，

定义\(f(x, y)\)为：

\[\begin{eqnarray*}
\prod\limits_{p \in prime(x)} g(y, p)
\end{eqnarray*}
\]

现在给出\(x, n\)，要求计算：

\[\prod\limits_{i = 1}^n f(x, i) \bmod (10^9 + 7)
\]

思路：

枚举\(x\)的每个质因子，再从高到低枚举每个质因子的幂次，考虑对于一个质因子\(p\)，用\(f[i]\)表示\([1, n]\)中有多少个\(p^i\)的倍数，且不是\(p^j (j > i)\)的倍数，那么个数是\(\left\lfloor n / p^i \right\rfloor - \left\lfloor n / p^{i + 1} \right\rfloor\)

代码：

view code

#include <bits/stdc++.h>

using namespace std;

#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }

#define fi first

#define se second

#define endl "\n"

using ll = long long;

using pII = pair <int, int>;

using pLL = pair <ll, ll>;

using VI = vector <int>;

using VL = vector <ll>;

constexpr int mod = 1e9 + 7;

template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }

template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }

template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }

inline int rd() { int x; cin >> x; return x; }

template <class T> inline void rd(T &x) { cin >> x; }

template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }

template <class T> inline void out(T s) { cout << s << "\n"; }

template <class T> inline void out(vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }

inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }

constexpr int N = 1e5 + 10;

ll x, n, bit[110];

inline ll ceil(ll x, ll y) {

	return (x + y - 1) / y;

}

void run() {

	vector <int> fac;

	for (ll i = 2; i * i <= x; ++i) {

		if (x % i == 0) fac.push_back(i);

		while (x % i == 0) x /= i;

	}

	if (x != 1) fac.push_back(x);

	ll res = 1;

	for (auto &it : fac) {

		if (it > n) continue;

		int k = 1; bit[1] = it;

		for (int i = 2; ; ++i) {

			if (bit[i - 1] > ceil(n, it)) {

				k = i - 1;

				break;

			}

			bit[i] = bit[i - 1] * it;

		}

		ll tot = 0;

		for (int i = k; i >= 1; --i) {

			ll p = n / bit[i];

			p -= tot;

			res = res * qpow(bit[i] % mod, p % (mod - 1)) % mod;

			tot += p;

		}

	}

	out(res);

}

int main() {

	ios::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);

	cout << fixed << setprecision(20);

	while (cin >> x >> n) run();

	return 0;

}

D. Complete Tripartite

题意：

现在给出一张\(n\)个点\(m\)条边的无向图，没有自环和重边，问能否将点分成三个集合，使得集合内部的点之间没有边相连，但任意两个点(他们分属不同的集合)有边相连。

如果可以，输出方案。

思路：

考虑同一点集里所有的点连出去的边都是相同的，那么根据这个\(Hash\)，如果刚好有三种\(Hash\)值，那么按\(Hash\)值分类即可

代码：

view code

#include <bits/stdc++.h>

using namespace std;

#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }

#define fi first

#define se second

#define endl "\n"

using ll = long long;

using ull = unsigned long long;

using pII = pair <int, int>;

using pLL = pair <ll, ll>;

using VI = vector <int>;

using VL = vector <ll>;

constexpr int mod = 1e9 + 7;

template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }

template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }

template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }

inline int rd() { int x; cin >> x; return x; }

template <class T> inline void rd(T &x) { cin >> x; }

template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }

template <class T> inline void out(T s) { cout << s << "\n"; }

template <class T> inline void out(vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }

inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }

constexpr int N = 3e5 + 10;

int n, m, ans[N];

mt19937 rnd(time(0));

ull f[N], g[N];

map <ull, vector<int>> mp;

void run() {

	for (int i = 1; i <= n; ++i) f[i] = rnd();

	memset(g, 0, sizeof g);

	mp.clear();

	for (int i = 1, u, v; i <= m; ++i) {

		cin >> u >> v;

		g[u] ^= f[v];

		g[v] ^= f[u];

	}

	for (int i = 1; i <= n; ++i) {

		mp[g[i]].push_back(i);

		if (mp.size() > 3) return out(-1);

	}

	if (mp.size() != 3) return out(-1);

	int cnt = 0;

	for (auto &it : mp) {

		++cnt;

		for (auto &u : it.second) {

			ans[u] = cnt;

		}

	}

	for (int i = 1; i <= n; ++i)

		cout << ans[i] << " \n"[i == n];

}

int main() {

	ios::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);

	cout << fixed << setprecision(20);

	while (cin >> n >> m) run();

	return 0;

}

E. Another Filling the Grid

题意：

给出一个\(n \cdot n\)的矩形，每个位置可以填\([1, k]\)。

现在要求每一行至少有一个\(1\)，每一列至少有一个\(1\)，问填数的方案数。

思路一：

考虑\(f[i][j]\)表示考虑前\(i\)行有\(j\)列有\(1\)，转移的时候注意每一行至少有一个\(1\)。

时间复杂度\(O(n^3)\)

代码一：

view code

#include <bits/stdc++.h>

using namespace std;

#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }

#define fi first

#define se second

#define endl "\n"

using ll = long long;

using pII = pair <int, int>;

using pLL = pair <ll, ll>;

using VI = vector <int>;

using VL = vector <ll>;

constexpr int mod = 1e9 + 7;

template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }

template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }

template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }

inline int rd() { int x; cin >> x; return x; }

template <class T> inline void rd(T &x) { cin >> x; }

template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }

template <class T> inline void out(T s) { cout << s << "\n"; }

template <class T> inline void out(vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }

inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }

constexpr int N = 300 + 10;

int n, K; ll f[N][N], C[N][N];

void run() {

	if (n == 1 || K == 1) return out(1);

	memset(f, 0, sizeof f);

	for (int i = 1; i <= n; ++i) {

		f[1][i] = C[n][i] * qpow(K - 1, n - i) % mod;

	}

	for (int i = 2; i <= n; ++i) {

		for (int j = 1; j <= n; ++j) {

			ll p = qpow(K, j);

			for (int k = j; k <= n; ++k) {

				if (k == j) chadd(f[i][k], f[i - 1][j] * (p + mod - qpow(K - 1, j)) % mod * qpow(K - 1, n - k) % mod);

				else chadd(f[i][k], f[i - 1][j] * p % mod * C[n - j][k - j] % mod * qpow(K - 1, n - k) % mod);

			}

		}

	}

	out(f[n][n]);

}

int main() {

	memset(C, 0, sizeof C);

	C[0][0] = 1;

	for (int i = 1; i < N; ++i) {

		C[i][0] = C[i][i] = 1;

		for (int j = 1; j < i; ++j)

			C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;

	}

	ios::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);

	cout << fixed << setprecision(20);

	while (cin >> n >> K) run();

	return 0;

}

思路二：

考虑枚举有\(i\)行\(j\)列没有\(1\)，然后根据\((i + j)\)的奇偶性容斥。

时间复杂度\(O(n^2)\)

代码二：

view code

#include <bits/stdc++.h>

using namespace std;

#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }

#define fi first

#define se second

#define endl "\n"

using ll = long long;

using pII = pair <int, int>;

using pLL = pair <ll, ll>;

using VI = vector <int>;

using VL = vector <ll>;

constexpr int mod = 1e9 + 7;

template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }

template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }

template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }

inline int rd() { int x; cin >> x; return x; }

template <class T> inline void rd(T &x) { cin >> x; }

template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }

template <class T> inline void out(T s) { cout << s << "\n"; }

template <class T> inline void out(vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }

inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }

constexpr int N = 300 + 10;

int n, K; ll f[N][N], C[N][N];

void run() {

	if (n == 1 || K == 1) return out(1);

	ll ans = 0;

	//枚举有i行，j列没有1，容斥

	for (int i = 0; i <= n; ++i) {

		for (int j = 0; j <= n; ++j) {

			ll ch = i * n + j * n - i * j;

			ll ex = n * n - ch;

			ll now = C[n][i] * C[n][j] % mod * qpow(K - 1, ch) % mod * qpow(K, ex) % mod;

			if ((i + j) & 1) chadd(ans, mod - now);

			else chadd(ans, now);

		}

	}

	out(ans);

}

int main() {

	C[0][0] = 1;

	for (int i = 1; i < N; ++i) {

		C[i][0] = C[i][i] = 1;

		for (int j = 1; j < i; ++j)

			C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;

	}

	ios::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);

	cout << fixed << setprecision(20);

	while (cin >> n >> K) run();

	return 0;

}

思路三：

考虑枚举至少有\(i\)行没有\(1\)，那么保证每一列都至少有一个\(1\)，那么答案就是：

\[\begin{eqnarray*}
\sum\limits_{i = 0}^{n - 1} (-1)^i{n \choose i}(k - 1)^{in}f^n(n - i)
\end{eqnarray*}
\]

其中\(f[i]\)表示有\(i\)个数，至少有一个\(1\)的方案数，显然有：

\[\begin{eqnarray*}
f[i] = k^{i} - (k - 1)^i
\end{eqnarray*}
\]

时间复杂度\(O(nlogk)\)

代码三：

view code

#include <bits/stdc++.h>

using namespace std;

#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }

#define fi first

#define se second

#define endl "\n"

using ll = long long;

using pII = pair <int, int>;

using pLL = pair <ll, ll>;

using VI = vector <int>;

using VL = vector <ll>;

constexpr int mod = 1e9 + 7;

template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }

template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }

template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }

inline int rd() { int x; cin >> x; return x; }

template <class T> inline void rd(T &x) { cin >> x; }

template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }

template <class T> inline void out(T s) { cout << s << "\n"; }

template <class T> inline void out(vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }

inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }

constexpr int N = 300 + 10;

int n, K; ll f[N], C[N][N];

void run() {

	if (n == 1 || K == 1) return out(1);

	ll ans = 0;

	//枚举有i行没有1，然后保证每列至少有一个1，容斥

	for (int i = 0; i < n; ++i) {

		ll f = (qpow(K, n - i) - qpow(K - 1, n - i) + mod) % mod;

		ll now = qpow(f, n) * C[n][i] % mod * qpow(K - 1, n * i) % mod;

		if (i & 1) chadd(ans, -now);

		else chadd(ans, now);

	}

	out(ans);

}

int main() {

	C[0][0] = 1;

	for (int i = 1; i < N; ++i) {

		C[i][0] = C[i][i] = 1;

		for (int j = 1; j < i; ++j)

			C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;

	}

	ios::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);

	cout << fixed << setprecision(20);

	while (cin >> n >> K) run();

	return 0;

}