Codeforces Round #589 (Div. 2)
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:
\prod\limits_{p \in prime(x)} g(y, p)
\end{eqnarray*}
\]
现在给出\(x, n\),要求计算:
\]
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:
\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:
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;
}
Codeforces Round #589 (Div. 2)的更多相关文章
- Codeforces Round #589 (Div. 2)-E. Another Filling the Grid-容斥定理
Codeforces Round #589 (Div. 2)-E. Another Filling the Grid-容斥定理 [Problem Description] 在\(n\times n\) ...
- Codeforces Round 589 (Div. 2) 题解
Is that a kind of fetishism? No, he is objectively a god. 见识了一把 Mcdic 究竟出题有多神. (虽然感觉还是吹过头了) 开了场 Virt ...
- Codeforces Round #589 (Div. 2) E. Another Filling the Grid(DP, 组合数学)
链接: https://codeforces.com/contest/1228/problem/E 题意: You have n×n square grid and an integer k. Put ...
- Codeforces Round #589 (Div. 2) D. Complete Tripartite(染色)
链接: https://codeforces.com/contest/1228/problem/D 题意: You have a simple undirected graph consisting ...
- Codeforces Round #589 (Div. 2) C - Primes and Multiplication(数学, 质数)
链接: https://codeforces.com/contest/1228/problem/C 题意: Let's introduce some definitions that will be ...
- Codeforces Round #589 (Div. 2) B. Filling the Grid
链接: https://codeforces.com/contest/1228/problem/B 题意: Suppose there is a h×w grid consisting of empt ...
- Codeforces Round #589 (Div. 2) A. Distinct Digits
链接: https://codeforces.com/contest/1228/problem/A 题意: You have two integers l and r. Find an integer ...
- Codeforces Round #589 (Div. 2) (e、f没写)
https://codeforces.com/contest/1228/problem/A A. Distinct Digits 超级简单嘻嘻,给你一个l和r然后寻找一个数,这个数要满足的条件是它的每 ...
- 【Codeforces Round #589 (Div. 2) D】Complete Tripartite
[链接] 我是链接,点我呀:) [题意] 题意 [题解] 其实这道题感觉有点狗. 思路大概是这样 先让所有的点都在1集合中. 然后随便选一个点x,访问它的出度y 显然tag[y]=2 因为和他相连了嘛 ...
随机推荐
- Java基础IO类之数据流
DataInputStream: 数据输入流允许应用程序以与机器无关方式从底层输入流中读取基本java数据类型.应用程序可以使用数据输出流 写入稍后由数据输入流读取的数据.DataInputStrea ...
- 全栈项目|小书架|服务器开发-Koa全局路由实现
什么是路由 路由就是具体的访问路径,指向特定的功能模块.一个api接口是由ip(域名)+端口号+路径组成,例如 :https://www.npmjs.com/package/koa-router就是一 ...
- Spring Cloud Alibaba学习笔记(4) - Feign配置与使用
什么是Feign Feign是一个声明式Web Service客户端. 使用Feign能让编写Web Service客户端更加简单, 它的使用方法是定义一个接口,然后在上面添加注解,同时也支持JAX- ...
- Presto个人常用操作
时间戳转日期: from_unixtime(1569168000,'yyyy-MM-dd') = '2019-09-23' '20190903'转为'2019-09-23': unix_timesta ...
- vue动态绘制四分之三圆环
参照网上的一个案例“参照的为绘制的是一个动态的圆环”,现在我的需求是改编成四分之三的圆环实现效果: 样式展示 canvas绘图基本操作设置就可以参考源代码链接:原文:https://blog.csdn ...
- 安卓开发之cache 的使用(图片查看器案例)
package com.lidaochen.test; import android.graphics.Bitmap; import android.graphics.BitmapFactory; i ...
- csv注入复现代码
以下代码生成的csv文件,使用Microsoft Execl能成功弹出计算器,虽然打开时有安全提示,但是大多数src还是会接收该类漏洞 -------------------------------- ...
- stm32内联汇编
首先,先看一下mdk下的混合编程的基本方法: 使用如上方法就可以进行混合编程了. 但是要特殊注意一点,个人感觉这个是直接调用一个代码段,并非一个函数,因为他不会保护调用这个代码段之前的现场.比如: 在 ...
- 记一次B类地址子网划分
说明:一般内网地址,没有要求严格的划分之类的.需要按要求严格划分一般都是公网地址 温馨提醒:由于笔者功底不是很深厚,只能说我们保证把数算对用相对简单的方法 现有一个公网ip,B类地址130.3.0.0 ...
- c# 属性成员