Codeforces Round #589 (Div. 2)

Posted 2020-11-15 dup4

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

Solved	A	B	C	D	E	F
5/6	O	O	O	O	?	-

• O 在比赛中通过
• ? 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;
}