2021-07-15 集训题解

光影交错

题目传送门

Description

Solution

可以看出，假设我们设置一个临界点 (n) ，当 (n) 足够大的时候，(n) 步操作之后对答案的影响就不再精度考虑范围之类了。

我们设 (f(i)) 分别表示 (i) 次操作时非中性灵气的期望出现次数，(g(i)) 表示 (i) 次非中性灵气是总体阳间的概率。

那么答案就是 (sum_{i=1}^{n} f(i) imes g(i)) 。

可以得到：

[g(x)=sum_{i=lfloorfrac{x}{2} floor+1}^{x} inom{x}{i}qL^iqR^{x-i} ]

我们可以得到递推式：

[left{egin{array}{l} g(x)=0, ext{when }x=0\ g(x)=g(x-1)-pDinom{x-1}{lfloorfrac{x}{2} floor} pL^{x/2}pD^{x/2-1}, ext{when }xequiv 0pmod{2}\ g(x)=g(x-1)+pLinom{x-1}{lfloorfrac{x}{2} floor} pL^{(x-1)/2}pD^{(x-1)/2}, ext{when }xequiv 1pmod{2} end{array} ight. ]

考虑如何求解 (f(x))，可以考虑对其构造生成函数 (F(x)) 。那么我们就有：

[F(x)=sum_{i=0}^{infty} f(i) imes x^i=sum_{i=1}^{infty} (1-p)^{i-1}((pL+pD)x+(1-pL-pD))^i=frac{1-p}{1-(1-p)[(pL+pD)x+(1-pL-pD)]} ]

然后你就可以得到：

[f(x)=frac{f(x-1)(1-p)(pL+pD)}{1-(1-p)(1-pL-pD)} ]

那么你就可以 (Theta(n)) 解决这个问题，(n) 大概是 (1e7) 左右。

update

这里提供 神·tly 的一种精确求值 (Theta(1)) 做法。

可以写出答案就是：

[sum_{l>r,e} inom{l+r+e}{l+r}inom{l+r}{l}pL^lpD^r(1-pL-pD)^e(1-p)^{l+r+e-1} ]

考虑提出 (e)，那么答案就是：

[frac{1}{1-p}sum_{l>r} inom{l+r}{l}(pL imes (1-p))^l(pD imes (1-p))^rsum_{e} inom{l+r+e}{e}(1-pL-pD)^e(1-p) ]

你发现 (e) 的贡献只与 (l+r) 有关，考虑它的生成函数，然后将 (pL imes (1-p))写成 (pL^{'})，(pD imes (1-p)) 写成 (pD^{'})，发现答案可以写成：

[sum_{l>r} inom{l+r}{l}(pL^{'})^l(pD^{'})^rfrac{1}{(1-pE)^{l+r+1}} ]

其中 (pE=(1-pL-pD) imes (1-p))，将 (pL^{'} imes (1-pE)) 写成 (pL^{''})，同理替换成 (pD^{''})，那么式子就是：

[frac{1}{(1-p)(1-pE)}sum_{l>r} inom{l+r}{l}(pL^{''})^l(pD^{''})^r ]

考虑提出 (pD^{''})，那么它的贡献就是：

[S_D(n)=sum_{i=0}^{n-1}(pD^{''})^iinom{n+i}{n} ]

(n) 也就是原式中的 (l)。

你发现通过错位相减可以得到：

[S_D(n)(1-pD^{''})=S_D(n-1)+inom{2(n-1)}{n-1}(pD^{''})^{n-1}-inom{2n-1}{n}(pD^{''})^n ]

为了方便，我们可以将 (inom{2(n-1)}{n-1}(pD^{''})^{n-1}-inom{2n-1}{n}(pD^{''})^n) 设为 (a_n)，那么 (r) 的贡献就可以表示为：

[sum_{i=1}^{l} frac{a_i}{(1-pD^{''})^{l-i+1}} ]

考虑交换求和符号，答案就是（忽略了前面的常数）：

[sum_{i=1}a_i imes (1-pD^{''})^{i-1}sum_{l=i}^{infty} (frac{pL^{''}}{1-pD^{''}})^l ]

将 (frac{pL^{''}}{1-pD^{''}}) 设为 (pH)，你发现答案就是：

[sum_{i=1} a_i imes (1-pD^{''})^{i-1}frac{pH^i}{1-pH} ]

[=frac{1}{(1-pH)(1-pD^{''})}sum_{i=1} a_i imes pL^{'} ]

考虑将 (a_i) 展开，你发现就是（忽略系数）：

[pL^{''}sum_{i=0}^{infty} inom{2i}{i}(pD^{''})^i(pL^{''})^{i}-sum_{i=1}^{infty} inom{2i-1}{i}(pD^{''})^i(pL^{''})^i ]

对于前面一个你发现假设 Catalan 数列的生成函数为 (C(x)) ，那么前面一个的生成函数（设 (x=pD^{''}pL^{''})）就是 ((C(x) imes x)^{'})，而我们知道 (C(x)=C^2(x) imes x+1)，所以前面一个的贡献就是 (qL^{''}frac{1}{sqrt{1-4x}})，后面一个类似，可以推出就是 (frac{1}{2x}(frac{1}{sqrt{1-4x}}-1))。

至此，你就可以做到 (Theta(1)) 计算了。

Code for O(n)

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define MAXN 10000005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c != ' ' && c != '
') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void chkmax (T &a,T b){a = max (a,b);}
template <typename T> void chkmin (T &a,T b){a = min (a,b);}

#define eps 1e-10
double p,pl,pd,f[MAXN],g[MAXN],drw[MAXN],tmp[MAXN];

void work (){
	scanf ("%lf%lf%lf",&pl,&pd,&p);
	double now = 1,ans = 0;int n = 0;tmp[0] = 1,tmp[1] = f[0] = f[1] = 0;
	while (now > eps){
		for (Int i = 1;~i;-- i){
			tmp[i] *= (1 - pl - pd);
			if (i) tmp[i] += tmp[i - 1] * (pl + pd);
			f[i] += tmp[i] * now;
		}
		now *= (1 - p),n ++;
	}
	double h = 1 - (1 - p) * (1 - pl - pd);
	for (Int i = 2;i <= n;++ i) f[i] = f[i - 1] * (1 - p) * (pl + pd) / h;
	double sum = pl + pd;pl /= sum,pd /= sum;
	for (Int i = 0;i <= n;++ i){
		if (i & 1){
			g[i] = g[i - 1] + drw[i - 1] * pl;
			if (i == 1) drw[i] = pl;
			else drw[i] = drw[i - 2] * pl * pd * i * (i - 1) / (i / 2) / (i / 2 + 1);
			ans += g[i] * f[i];
		}
		else{
			if (i == 0) drw[i] = 1;
			else drw[i] = 4 * drw[i - 2] * (i - 1) / i * pl * pd;
			if (i) g[i] = g[i - 1] - drw[i - 1] * pd;
			ans += g[i] * f[i];
		}
	}
	printf ("%.10f
",ans);
}

signed main(){
	freopen ("augury.in","r",stdin);
	freopen ("augury.out","w",stdout);
	int num,T;read (num,T);
	while (T --> 0) work ();
	return 0;
}

Code for O(1)

#include <bits/stdc++.h>
using namespace std;

#define Int register int

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c != ' ' && c != '
') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void chkmax (T &a,T b){a = max (a,b);}
template <typename T> void chkmin (T &a,T b){a = min (a,b);}

double pL,pD,p;
void Work (){
	scanf ("%lf%lf%lf",&pL,&pD,&p);
	if (p == 1) return printf ("%.10f
",pL),void ();
	if (pL == 0) return printf ("%.10f
",0.0),void();
	double pE = (1 - pL - pD) * (1 - p),pL2 = pL * (1 - p) / (1 - pE),pD2 = pD * (1 - p) / (1 - pE);
	double pH = pL2 / (1 - pD2),r = pD2 * pL2,u = 1.0 / sqrt(1 - 4 * r);
	printf ("%.10f
",pL2 / (1 - p) / (1 - pE) / (1 - pH) / (1 - pD2) * (!pD2 || !pL2 ? 1 : u - pD2 / (2 * r) * (u - 1)));
}

signed main(){
	freopen ("augury.in","r",stdin);
	freopen ("augury.out","w",stdout);
	int num,T;read (num);read (T);
	while (T --> 0) Work ();
	return 0;
}

七星连珠

题目传送门

Description

Solution

md，处理 (k) 进制 ( ext{FWT}) 根本不是难点好吧。。。

考虑 (2) 进制时如何处理，可以发现，如果我们如果对于每个位置 ((i,j))，构造一个数组，并将 (a_{i,j}) 赋值，再做 ( ext{DWT})，对整个矩阵求个行列式（可以看出这是一个 ( ext{DWT}) 数组），拿行列式再 ( ext{IDWT}) 回去，有值的位置就很有可能是正确答案。为什么说很有可能呢？因为行列式要乘 ((-1)^{pi(p)})，就有可能把本来有值的消成 (0)，所以我们将 (a_{i,j}) 位置赋值的时候需要赋上一个随机值。

至于 (3) 进制，随便高维 ( ext{FWT}) 搞搞就行了。因为要求单位根，所以找个模数 (10^9+9)，原根就是 (13)。

Code

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define mod 1000000009
#define MAXN 55

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void chkmax (T &a,T b){a = max (a,b);}
template <typename T> void chkmin (T &a,T b){a = min (a,b);}

int N,K,w[3][3],val[MAXN][MAXN];
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int b){
	int res = 1;for (;b;b >>= 1,a = mul (a,a)) if (b & 1) res = mul (res,a);
	return res;
}
int inv (int x){return qkpow (x,mod - 2);}
void Add (int &a,int b){a = add (a,b);}
void Sub (int &a,int b){a = dec (a,b);}

#define MAXM 2187

int ivk,lim;
struct node{//储存一个fwt数组
	int a[MAXM];
	node(){memset (a,0,sizeof (a));}
	int & operator [](const int key){return a[key];}
	void fwt (int type){
		for (Int i = 1;i < lim;i *= K)
			for (Int j = 0;j < lim;j += i * K)
				for (Int k = 0;k < i;++ k){
					int tx[3] = {};
					for (Int d = 0;d < K;++ d) for (Int e = 0;e < K;++ e) Add (tx[d],mul (type == 1 ? w[d][e] : inv(w[d][e]),a[j + k + i * e]));
					for (Int d = 0;d < K;++ d) a[j + k + i * d] = mul (tx[d],type == 1 ? 1 : ivk);
				}
	}
}t,fuc[MAXN][MAXN];

int mat[MAXN][MAXN];
int getdet (){
	int ans = 1;
	for (Int i = 1;i <= N;++ i){
		int pos = i;
		for (Int j = i + 1;j <= N;++ j) if (mat[j][i]){pos = j;break;}
		if (pos != i) swap (mat[i],mat[pos]),ans = mod - ans;
		for (Int j = i + 1,iv = qkpow (mat[i][i],mod - 2);j <= N;++ j){
			int del = mul (mat[j][i],iv);
			for (Int k = i;k <= N;++ k) Sub (mat[j][k],mul (mat[i][k],del));
		}
		ans = mul (ans,mat[i][i]);
	}
	return ans;
}

signed main(){
	freopen ("astrology.in","r",stdin);
	freopen ("astrology.out","w",stdout);
	srand (time(NULL));
	int num;read (num,N,K),ivk = qkpow (K,mod - 2),lim = qkpow (K,7);
	for (Int i = 1;i <= N;++ i)
		for (Int j = 1;j <= N;++ j)
			read (val[i][j]);
	if (K == 2) w[0][0] = w[0][1] = w[1][0] = 1,w[1][1] = mod - 1;
	else{
		int gi = qkpow (13,(mod - 1) / 3);
		for (Int i = 0;i < 3;++ i) w[i][0] = 1,w[i][1] = qkpow (gi,i),w[i][2] = mul (w[i][1],w[i][1]);
	}
	for (Int i = 1;i <= N;++ i)
		for (Int j = 1;j <= N;++ j)
			fuc[i][j][val[i][j]] = rand(),fuc[i][j].fwt(1);
	for (Int x = 0;x < lim;++ x){
		for (Int i = 1;i <= N;++ i)
			for (Int j = 1;j <= N;++ j)
				mat[i][j] = fuc[i][j][x];
		t[x] = getdet();
	}
	t.fwt(-1);
	for (Int i = 0;i < lim;++ i) if (t[i]) write (i),putchar (' ');
	putchar ('
');
	return 0;
}

魔法咒语

题目传送门

Description

Solution

可以发现，答案就是：

[sum_{h=0}^{N-1} f(h)sum_{n=h+1}^{N} (-1)^{n-h-1}inom{n}{h} ]

其中，(f(h)=sum_{j=0}^{k} h^j)。

考虑后面那个东西如何递推。

因为存在 (inom{h}{i}=inom{h-1}{i-1}+inom{h-1}{i})，通过移项我们可以得到 (inom{h-1}{i-1}=inom{h}{i}-inom{h-1}{i}Rightarrow inom{h}{i}=inom{h+1}{i+1}-inom{h}{i+1}) 。

将其代入原式中，我们可以得到：

[g(i)=sum_{h=i}^{N} (-1)^{h-i}inom{h}{i} ]

[=sum_{h=i}^{n} (-1)^{h-i}inom{h+1}{i+1}+sum_{h=i}^{n} (-1)^{h-i-1}inom{h}{i+1} ]

[=2 imes g_{i+1}+(-1)^{N-i}inom{N+1}{i+1} ]

然后你发现 (i^{k+1}) 用线性筛筛出来就好了，复杂度 (Theta(n))。

Code

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define int long long
#define MAXN 10000005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c != ' ' && c != '
') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> void chkmax (T &a,T b){a = max (a,b);}
template <typename T> void chkmin (T &a,T b){a = min (a,b);}

bool vis[MAXN];
int N,k,tot,mod,g[MAXN],pw[MAXN],fac[MAXN],ifac[MAXN],prime[MAXN];

int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int b){
	b %= (mod - 1),a %= mod;int res = 1;for (;b;b >>= 1,a = mul (a,a)) if (b & 1) res = mul(res,a);
	return res;
}
void Add (int &a,int b){a = add (a,b);}
void Sub (int &a,int b){a = dec (a,b);}

void Euler (int up){
	pw[1] = 1;
	for (Int i = 2;i <= up;++ i){
		if (!vis[i]) prime[++ tot] = i,pw[i] = qkpow (i,k + 1);
		for (Int j = 1;j <= tot && i * prime[j] <= up;++ j){
			vis[i * prime[j]] = 1,pw[i * prime[j]] = mul (pw[i],pw[prime[j]]);
			if (i % prime[j] == 0) break;
		}
	}
}

int inv (int x){return mul (ifac[x],fac[x - 1]);}
int binom (int a,int b){return a >= b ? mul (fac[a],mul (ifac[b],ifac[a - b])) : 0;}
int f (int x){return x == 0 ? 1 : (x == 1 ? (k + 1) % mod : mul (pw[x] - 1,inv (x - 1)));}

signed main(){
	freopen ("abracadabra.in","r",stdin);
	freopen ("abracadabra.out","w",stdout);
	int num;read (num,N,k,mod),Euler (N);
	fac[0] = 1;for (Int i = 1;i <= N + 1;++ i) fac[i] = mul (fac[i - 1],i);
	ifac[N + 1] = qkpow (fac[N + 1],mod - 2);for (Int i = N + 1;i;-- i) ifac[i - 1] = mul (ifac[i],i);
	for (Int i = N;i >= 0;-- i) g[i] = add (add (g[i + 1],g[i + 1]),mul (N - i & 1 ? mod - 1 : 1,binom (N + 1,i + 1)));
	int ans = 0;for (Int h = 0;h <= N;++ h) Add (ans,f(h)),Sub (ans,mul (f(h),g[h]));
	write (ans),putchar ('
');
	return 0;
}