이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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());
}
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