답안 #19624

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
19624	2016-02-25T02:32:34 Z	kaTkaHr	카드 (kriii4_Z)	C++14	0 / 100	1000 ms	142448 KB

Z

#include <stdio.h>
#include<vector>
#include <algorithm>
#include <map>

using namespace std;

typedef long long ll;

const int MX = 3005, MM = 1000000007;

template<typename T>
T pw(T A, ll B){
  T R = 1;
  while(B){
    if( B&1 ) R = R * A;
    A = A * A; B /= 2;
  }
  return R;
}

ll rv(ll A){ 
  ll R = 1, B = MM-2;
  while(B){
    if( B&1 ) R = R * A % MM;
    A = A * A % MM; B /= 2;
  }
  return R;
}
/*
struct frac{
  ll A, B;
		frac(ll A):A(A), B(1){}
  frac(ll a, ll b){
    A = (a%MM+MM) % MM;
    B = (b%MM+MM) % MM;
  }
  frac(){A = 0, B = 1;}
  frac operator+ (const frac &l)const{
    return frac((A * l.B + B * l.A) % MM, B * l.B % MM);
  }
  frac operator*(const frac &l)const{
    return frac(A*l.A % MM, B*l.B % MM);
  }
  frac operator/(const frac &l)const{
    return frac(A*l.B % MM, B*l.A % MM);
  }
  frac operator- (const frac &l)const{
    return frac((A*l.B - B*l.A%MM + MM) % MM, B*l.B % MM);
  }
  ll v(){ return A * rv(B) % MM; }
};// */
//*
struct frac{
  ll A;
		frac(ll A):A((A%MM+MM)%MM){}
  frac(ll a, ll b){
			a = (a%MM+MM)%MM;
			b = (b%MM+MM)%MM;
			A = rv(b) * a % MM;
  }
  frac(){A = 0;}
  frac operator+ (const frac &l)const{
			return l.A + A >= MM? l.A + A - MM: l.A + A;
  }
  frac operator*(const frac &l)const{
			return l.A * A % MM;
  }
  frac operator/(const frac &l)const{
			return A * rv(l.A) % MM;
  }
  frac operator- (const frac &l)const{
			return A >= l.A? A-l.A: A-l.A + MM;
  }
  ll v(){ return A; }
};// */

int D[MX];
frac T[MX][MX];
frac C[MX][MX];
frac A[11][MX];

int main()
{
	int N, L;
	scanf("%d%d", &N, &L);
	for(int i = 1; i <= N; i++){
		int A;
		scanf("%d", &A);
		D[A]++;
	}
	C[0][0] = 1;
	for(int i = 1; i <= 3000; i++){
		C[i][0] = 1;
		for(int j = 1; j <= i; j++){
			C[i][j] = C[i-1][j] + C[i-1][j-1];
		}
	}
	A[0][0] = 1;
	for(int i = 1; i <= L; i++) A[0][i] = A[0][i-1] * D[0];
	for(int k = 1; k <= 10; k++){
		T[0][0] = 1;
		for(int i = 1; i <= D[k]; i++){
			for(int j = 1; j <= L; j++){
				T[i][j] = T[i][j-1] * i;
				if( j >= k) T[i][j] = T[i][j] + T[i-1][j-k];
			}
		}
		frac mul = 1;
		for(int i = 1; i <= D[k]; i++) mul = mul * i;
		for(int i = D[k]*k; i <= L; i++){
			frac tmp = 0;
			for(int j = D[k]*k; j <= i; j++){
				tmp = tmp + A[k-1][i-j] * T[D[k]][j] * C[i][j];
			}
			A[k][i] = tmp;
		}
		for(int i = 0; i <= L; i++) A[k][i] = A[k][i] * mul;
	}

	frac ans = A[10][L];
	for(int i = 1; i <= L; i++) ans = ans / N;
	printf("%lld\n", ans.v());
}

Subtask #1

0.0 / 13.0

#	결과	실행 시간	메모리	Grader output
1	Execution timed out	1000 ms	142448 KB	Program timed out
2	Halted	0 ms	0 KB	-

Subtask #2

0.0 / 87.0

#	결과	실행 시간	메모리	Grader output
1	Halted	0 ms	0 KB	-