目录
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：
[egin{eqnarray*}
prodlimits_{p in prime(x)} g(y, p)
end{eqnarray*}
]
现在给出(x, n)，要求计算：
[prodlimits_{i = 1}^n f(x, i) mod (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(leftlfloor n / p^i
ight
floor - leftlfloor n / p^{i + 1}
ight
floor)
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 "
"
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 << "
"; }
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 "
"
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 << "
"; }
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] << "
"[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 "
"
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 << "
"; }
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 "
"
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 << "
"; }
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：
[egin{eqnarray*}
sumlimits_{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：
[egin{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 "
"
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 << "
"; }
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;
}