2023杭电多校第五场 - 1 7 12

目录

• 1001 Typhoon
• 1007 Expectation (Easy Version)
• 1012 Counting Stars

比赛地址：传送门
赛时过了 4 题，今天这场把这辈子的 TLE 都吃了一遍。。。
后期直接坐牢，还得加练啊！
1001 计算几何 点到线段的距离
1007 二项式定理求期望
1012 求组合数

---

1001 Typhoon

题意
给你一条折线路径，再给你一系列点，问你这些点到该折线的最短距离？

思路
计算几何基础题
本题模拟即可，对于每个点遍历所有折线，取距离最小值
$O(mn)$ 即可通过
没有计算几何板子怎么办？ 现在就开始准备！！！

代码

//>>>Qiansui
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x, y, sizeof(x))
#define debug(x) cout << #x << " = " << x << '\n'
#define debug2(x,y) cout << #x << " = " << x << " " << #y << " = "<< y << '\n'
//#define int long long
 
using namespace std;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pull;
typedef pair<double, double> pdd;
/*
 
*/
const int maxm = 2e5 + 5, inf = 0x3f3f3f3f, mod = 998244353;
 
const double pi = acos(-1.0);
const double eps = 1e-8;
 
int sgn(double x){//判断x的大小
	if(fabs(x)<eps) return 0;
	else return x<0?-1:1;
}
 
int dcmp(double x,double y){//比较两个浮点数
	return sgn(x - y);
}
 
inline int dcmp(double a){
	return a<-eps?-1:(a>eps?1:0); //处理精度
}
 
struct Point{  
	double x,y;
	Point(){}
	Point(double x,double y):x(x),y(y){}
 
	Point operator + (const Point& B){return Point(x+B.x,y+B.y);}
	Point operator - (const Point& B){return Point(x-B.x,y-B.y);}
};
 
typedef Point Vector;
 
double Dot(const Vector& A, const Vector& B){ return A.x*B.x + A.y*B.y; }
double Cross(const Vector& A, const Vector& B){return A.x*B.y - A.y*B.x;}
double Distance(Point A,Point B){ return sqrt((A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y)); }
 
struct Line{
	Point p1,p2;                  //直线上的两个点构成Line
	Line(){}
	Line(Point p1,Point p2):p1(p1),p2(p2){}
};
 
typedef Line Segment;
 
double Dis_point_line(Point &p, Line &v){
	return fabs(Cross(p-v.p1,v.p2-v.p1))/Distance(v.p1,v.p2);
}
 
double Dis_point_seg(Point &p, Segment &v){
	if(sgn(Dot(p- v.p1,v.p2-v.p1)) < 0 || sgn(Dot(p- v.p2,v.p1-v.p2)) < 0)
		return min(Distance(p,v.p1),Distance(p,v.p2));
	return Dis_point_line(p,v);           //点的投影在线段上
}
 
void solve(){
	int n, m;
	cin >> m >> n;
	Point a, b;
	vector<Line> l;
	for(int i = 0; i < m; ++ i){
		cin >> a.x >> a.y;
		if(i){
			l.push_back(Line(a,b));
			b = a;
		}else{
			b = a;
		}
	}
	int len = l.size();
	for(int i = 0; i < n; ++ i){
		cin >> a.x >> a.y;
		double ans;
		for(int j = 0; j < len; ++ j){
			if(j){
				double t = Dis_point_seg(a, l[j]);
				if(dcmp(ans, t) == 1) ans = t;
			}else ans = Dis_point_seg(a, l[j]);
		}
		cout << fixed << setprecision(4) << ans << '\n';
	}
	return ;
}
 
signed main(){
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	int _ = 1;
	// cin >> _;
	while(_ --){
		solve();
	}
	return 0;
}
 

---

1007 Expectation (Easy Version)

题意
你可以进行游戏 n 轮，每次游戏你赢的概率为 $\frac{a}{b}$。每次你赢的时候，你将会获得 $k^m$分， k 为你目前一共赢的次数。问你最后分数的期望取模 998244353

思路
由题可得 $P(X=i)=C_n^i·p^i·(1-p{)}^i$
故 $ans={\displaystyle \sum _{i=1}^{n}P(X=i)·{\displaystyle \sum _{j=1}^i{j}^m}}$
记得取模即可
赛时TLE....................... 优化是将 $p^i$ 和 $(1-p{)}^{n-i}$ 提取出来，每次遍历 k 的时候再单独成，而不用 qpow 来求，这样就不 T 了

代码

//>>>Qiansui
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x, y, sizeof(x))
#define debug(x) cout << #x << " = " << x << '\n'
#define debug2(x,y) cout << #x << " = " << x << " " << #y << " = "<< y << '\n'
//#define int long long
 
using namespace std;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pull;
typedef pair<double, double> pdd;
/*
 
*/
const int maxm = 2e5 + 5, inf = 0x3f3f3f3f;
 
const int N = 1e6 + 10, mod = 998244353;
ll fact[N], finv[N];
ll qpow(ll a, ll x){//带模快速幂
	ll res = 1;
	while(x){
		if(x & 1) res = res * a % mod;
		x >>= 1;
		a = a * a % mod;
	}
	return res;
}
 
void pre(){//线性预处理
	fact[0] = 1;
	for(int i = 1; i < N; ++ i){
		fact[i] = fact[i - 1] * i % mod;
	}
	finv[0] = 1;
	finv[N - 1] = qpow(fact[N - 1], mod - 2);
	for(int i = N - 2; i > 0; -- i)
		finv[i] = finv[i + 1] * (i + 1) % mod;
	return ;
}
 
ll C(int a, int b){//求取组合数
	if(a < 0 || b < 0) return 1;
	return fact[a] * finv[b] % mod * finv[a - b] % mod;
}
 
void solve(){
	int n, m, a, b;
	cin >> n >> m >> a >> b;
	ll invb = qpow(b, mod - 2), ans = 0, temp, sum = 0;
	ll p = a * invb % mod, q = (b - a) * invb % mod;
	ll pp = 1, qq = qpow(q, n), invq = qpow(q, mod - 2);
	for(int i = 1; i <= n; ++ i){
		sum = (sum + qpow(i, m)) % mod;
		pp = pp * p % mod;
		qq = qq * invq % mod;
		temp = C(n, i) * pp % mod * qq % mod * sum % mod;
		ans = (ans + temp) % mod;
	}
	cout << ans << '\n';
	return ;
}
 
signed main(){
	pre();
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	int _ = 1;
	cin >> _;
	while(_ --){
		solve();
	}
	return 0;
}
 

---

1012 Counting Stars

题意
给你一个 n 个顶点，m 条边的无向图，问你该图中存在的 k-star 图在 $k\in [2,n-1]$ 范围的个数取模 ${10}^9+7$ 后的异或和？

思路
暴力，对于每个顶点依次枚举其为中心，在其所连边中选 k 条边即为 k-star 图，之后求组合数统计每个 k 的个数，最后求异或和即可
赛时一直TLE....

代码

//>>>Qiansui
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x, y, sizeof(x))
#define debug(x) cout << #x << " = " << x << '\n'
#define debug2(x,y) cout << #x << " = " << x << " " << #y << " = "<< y << '\n'
//#define int long long
 
using namespace std;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pull;
typedef pair<double, double> pdd;
/*
 
*/
const int maxm = 2e5 + 5, inf = 0x3f3f3f3f;
 
const int N = 1e6 + 10, mod = 1e9 + 7;
ll fact[N], finv[N];
ll qpow(ll a, ll x){//带模快速幂
	ll res = 1;
	while(x){
		if(x & 1) res = res * a % mod;
		x >>= 1;
		a = a * a % mod;
	}
	return res;
}
 
void pre(){//线性预处理
	fact[0] = 1;
	for(int i = 1; i < N; ++ i){
		fact[i] = fact[i - 1] * i % mod;
	}
	finv[0] = 1;
	finv[N - 1] = qpow(fact[N - 1], mod - 2);
	for(int i = N - 2; i > 0; -- i)
		finv[i] = finv[i + 1] * (i + 1) % mod;
	return ;
}
 
ll C(int a, int b){//求取组合数
	if(a < 0 || b < 0) return 1;
	return fact[a] * finv[b] % mod * finv[a - b] % mod;
}
 
void solve(){
	int n, m;
	cin >> n >> m;
	vector<int> d(n + 1, 0);
	for(int i = 0; i < m; ++ i){
		int u, v;
		cin >> u >> v;
		++ d[u]; ++ d[v];
	}
	ll ans = 0;
	vector<int> cnt(n + 1, 0);
	for(int i = 1; i <= n; ++ i){
		if(d[i] >= 2){
			for(int j = 2; j <= min(n - 1, d[i]); ++ j){
				cnt[j] = (cnt[j] + C(d[i], j)) % mod;
			}
		}
	}
	for(int i = 2; i <= n - 1; ++ i){
		ans ^= cnt[i];
	}
	cout << ans << '\n';
	return ;
}
 
signed main(){
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	pre();
	int _ = 1;
	cin >> _;
	while(_ --){
		solve();
	}
	return 0;
}