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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#66642#5169. 夹娃娃Dual0 0ms0kbC++4.5kb2022-12-09 10:17:592022-12-09 10:18:02

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-12-09 10:18:02]
  • 评测
  • 测评结果:0
  • 用时:0ms
  • 内存:0kb
  • [2022-12-09 10:17:59]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
const int N = 17,V = 520,Lim = 7800; // V 和 Lim 都是是真实上界,没加 1 的那种

struct fastmod {
  typedef unsigned long long u64;
  typedef __uint128_t u128;

  int m;
  u64 b;

  fastmod(int m) : m(m), b(((u128)1 << 64) / m) {}
  int reduce(u64 a) {
    u64 q = ((u128)a * b) >> 64;
    int r = a - q * m;
    return r < m ? r : r - m;
  }
} Z(2);

int n,Q,P;
int a[N];
int b[N][V + 5],c[N][V + 5];
int F[N][V + 5]; // F_i 的系数表示

int fac[Lim + 5],ifac[Lim + 5],inv[Lim + 5];
int RLim;
inline int Add(const int &a,const int &b) { return (a + b >= P) ? (a + b - P) : (a + b);}
inline int Sub(const int &a,const int &b) { return (a < b) ? (a - b + P) : (a - b);}
inline void Plus(int &a,const int &b) { a += b;if(a >= P) a -= P;}
inline void Minus(int &a,const int &b) { a -= b;if(a < 0) a += P;}
inline int qpow(int a,int b) { int res = 1;while(b) {if(b&1) res = 1ll * res * a % P;a = 1ll * a * a % P;b >>= 1;} return res;}
inline void init(int n)
{
	fac[0] = 1;
	for(int i = 1;i <= n;i++) fac[i] = 1ll * fac[i - 1] * i % P;
	inv[1] = 1;
	for(int i = 2;i <= n;i++) inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
	ifac[0] = 1;
	for(int i = 1;i <= n;i++) ifac[i] = 1ll * ifac[i - 1] * inv[i] % P;
}
inline void Exp(int *f,int *g,int n)
{
	g[0] = 1;
	for(int i = 1;i <= n;i++)
	{
		int res = 0;
		for(int j = 1;j <= i;j++)
			res = Add(res,Z.reduce(1ll * Z.reduce(1ll * j * f[j]) * g[i - j]) );
		g[i] = Z.reduce(1ll * res * inv[i]);
	}
}
inline void Do_F(int id)
{
	static int lnf[V + 5];
	memset(lnf,0,sizeof lnf);
	for(int j = 1;j <= a[id];j++)
	{
		int tmp = b[id][j] * (c[id][j] + 1);
		for(int k = 1;tmp * k <= V;k++)
			// -x^{ij} / j
			Minus(lnf[tmp * k],inv[k]);
		tmp = b[id][j];
		for(int k = 1;tmp * k <= V;k++)
			Plus(lnf[tmp * k],inv[k]);
	}
	Exp(lnf,F[id],V);
}
int Ys[Lim + 5][N][V + 5]; // 前缀点值
int Pack[Lim + 5][Lim + 5];
inline void init_Largange(int n) // 1 - n + 1
{
	static int f[Lim + 5],tmp[Lim + 5];
	memset(f,0,(n + 1) << 2);
	f[0] = 1;
	for(int i = 1;i <= n + 1;i++)
	{
		for(int j = n + 1;j >= 1;j--)
			f[j] = Z.reduce(1ll * f[j] * (P - i)),Plus(f[j],f[j - 1]);
		f[0] = Z.reduce(1ll * f[0] * (P - i));
	}
	// printf("f: ");
	// for(int i = 0;i <= n + 1;i++) printf("%d ",f[i]);
	// printf("\n");
	for(int i = 1;i <= n + 1;i++)
	{
		memcpy(tmp,f,(n + 2) << 2);
		for(int j = 0;j <= n;j++)
			tmp[j] = P - Z.reduce(1ll * tmp[j] * inv[i]),Minus(tmp[j + 1],tmp[j]);
		int res = Z.reduce(1ll * ifac[i - 1] * ifac[n + 1 - i]);
		if((n + 1 - i) & 1) res = P - res;
		Pack[i][0] = Z.reduce(1ll * res * tmp[0]);
		for(int j = 1;j <= n;j++)
			Pack[i][j] = Add(Pack[i][j - 1] , Z.reduce(1ll * res * tmp[j]) );
	}
	// for(int j = 1;j <= n;j++) Plus(cst[j],cst[j - 1]);
}
inline int GetY(const string &S,int k,int X)
{
	int res = 1;
	for(int i = 1;i <= n;i++)
		if(S[i - 1] == '0')
			res = Z.reduce(1ll * res * Ys[X][i][0]);
		else res = Z.reduce(1ll * res * Ys[X][i][k]);
	return res;
}
inline int Solve(const string &S,int k,int m)
{
	int res = 0;
	for(int X = 1;X <= RLim + 1;X++)
		Plus(res,Z.reduce(1ll * GetY(S,k,X) * Pack[X][m]));
	return res;
}
int main()
{
	freopen("a.in","r",stdin);
	// freopen("log.out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin >> n >> Q >> P;
	Z = fastmod(P);
	RLim = V * n;
	for(int i = 1;i <= n;i++)
	{
		cin >> a[i];
		for(int j = 1;j <= a[i];j++)
			cin >> b[i][j] >> c[i][j];
	}
	init(RLim + 1);
	for(int i = 1;i <= n;i++)
		Do_F(i);
	// for(int i = 1;i <= n;i++)
	// {
	// 	printf("F[%d]:",i);
	// 	for(int j = 0;j <= 50;j++)
	// 		printf("%d ",F[i][j]);
	// 	printf("\n");
	// }
	// cout << "clock1 :" << 1.0 * clock() / CLOCKS_PER_SEC << endl;
	for(int X = 1;X <= RLim + 1;X++) // 插值 1 - Lim + 1,0 不知道可撤销背包的正确性
	{
		for(int i = 1;i <= n;i++)
		{
			int nowx = qpow(X,V);
			Ys[X][i][V] = Z.reduce(1ll * F[i][V] * nowx);
			for(int j = V - 1;j >= 0;j--)
			{			
				nowx = Z.reduce(1ll * nowx * inv[X]);
				Ys[X][i][j] = Add(Ys[X][i][j + 1],Z.reduce(1ll * nowx * F[i][j]));
				// printf("Ys[%d][%d][%d]=%d\n",X,i,j,Ys[X][i][j]);
			}
		}
	}
	// cout << "clock2 :" << 1.0 * clock() / CLOCKS_PER_SEC << endl;

	init_Largange(RLim);
	// cout << "clock3 :" << 1.0 * clock() / CLOCKS_PER_SEC << endl;

	while(Q--)
	{
		string S;
		int k,m;
		cin >> S >> m >> k;
		cout << Solve(S,k,m) << endl;
	}
	return 0;
}


详细

Subtask #1:

score: 0
Runtime Error

Test #1:

score: 0
Runtime Error

input:

1 521 998244353
39 520 520 11 22 414 8 95 18 229 356 26 407 316 10 24 26 19 61 11 130 482 476 420 15 192 193 208 24 19 233 494 217 275 294 26 28 439 20 272 277 28 198 5 335 22 8 28 17 154 78 6 13 175 17 2 5 477 256 200 4 1 36 427 371 439 23 10 65 426 25 24 27 121 29 28 13 12 453
0 520 1
1 519 1
1 51...

output:


result:


Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Skipped

Dependency #2:

0%

Subtask #4:

score: 0
Runtime Error

Test #9:

score: 0
Runtime Error

input:

15 52099 998244353
1 9 3
1 9 4
1 9 2
1 8 10
1 4 4
1 3 1
1 2 5
1 4 9
1 1 4
1 9 4
1 7 6
1 1 6
1 2 5
1 5 2
1 3 5
101000000001010 516 1
010001001010101 520 2
000000101000001 519 2
101011111100011 518 1
010110001000111 520 2
000110111100111 516 1
000100101001011 519 3
000111001010011 518 1
00001110010111...

output:


result:


Subtask #5:

score: 0
Skipped

Dependency #3:

0%

Subtask #6:

score: 0
Skipped

Dependency #4:

0%

Subtask #7:

score: 0
Skipped

Dependency #6:

0%