답안 #19645

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
19645	2016-02-25T03:20:51 Z	kaTkaHr	카드 (kriii4_Z)	C++14	13 / 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] * C[j-1][k-1];
			}
		}
		frac mul = 1;
		for(int i = 1; i <= D[k]; i++) mul = mul * i;
		if( D[k] ){
			for(int i = k*D[k]; i <= L; i++){
				for(int j = 0; j <= i; j++){
					A[k][i] = A[k][i] + A[k-1][j] * T[D[k]][i-j] * C[i][j];
				}
			}
		}
		else{
			for(int i = 0; i <= L; i++) A[k][i] = A[k-1][i];
		}
		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

13.0 / 13.0

#	결과	실행 시간	메모리	Grader output
1	Correct	262 ms	142448 KB	Output is correct
2	Correct	343 ms	142448 KB	Output is correct
3	Correct	281 ms	142448 KB	Output is correct
4	Correct	353 ms	142448 KB	Output is correct
5	Correct	282 ms	142448 KB	Output is correct
6	Correct	192 ms	142448 KB	Output is correct
7	Correct	292 ms	142448 KB	Output is correct
8	Correct	241 ms	142448 KB	Output is correct
9	Correct	357 ms	142448 KB	Output is correct
10	Correct	273 ms	142448 KB	Output is correct
11	Correct	304 ms	142448 KB	Output is correct
12	Correct	217 ms	142448 KB	Output is correct
13	Correct	257 ms	142448 KB	Output is correct
14	Correct	317 ms	142448 KB	Output is correct
15	Correct	326 ms	142448 KB	Output is correct
16	Correct	221 ms	142448 KB	Output is correct
17	Correct	297 ms	142448 KB	Output is correct
18	Correct	262 ms	142448 KB	Output is correct
19	Correct	204 ms	142448 KB	Output is correct
20	Correct	293 ms	142448 KB	Output is correct
21	Correct	397 ms	142448 KB	Output is correct
22	Correct	407 ms	142448 KB	Output is correct
23	Correct	406 ms	142448 KB	Output is correct
24	Correct	401 ms	142448 KB	Output is correct
25	Correct	394 ms	142448 KB	Output is correct

Subtask #2

0.0 / 87.0

#	결과	실행 시간	메모리	Grader output
1	Correct	571 ms	142448 KB	Output is correct
2	Correct	653 ms	142448 KB	Output is correct
3	Execution timed out	1000 ms	142448 KB	Program timed out
4	Halted	0 ms	0 KB	-