Codeforces Round #589 (Div. 2)

目录

• Contest Info
• Solutions
  • A. Distinct Digits
  • B. Filling the Grid
  • C. Primes and Multiplication
  • D. Complete Tripartite
  • E. Another Filling the Grid

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;
}