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 <string.h>
#include <ctype.h>
#include <time.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <numeric>
#include <functional>
#define MOD 1000000007
#define MAX 0x3f3f3f3f
#define MAX2 0x3f3f3f3f3f3f3f3fll
#define ERR 1e-10
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#pragma warning(disable:4996)
using namespace std;
typedef long long ll;
typedef long double ldb;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
ll mul_inv(ll a, ll b=MOD)
{
ll b0=b, t, q;
ll x0=0, x1=1;
if(b == 1) return 1;
while(a > 1)
{
q=a/b;
t=b, b=a%b, a=t;
t=x0, x0=x1-q*x0, x1=t;
}
if(x1 < 0) x1+=b0;
return x1;
}
ll modpow(ll x, ll y)
{
ll ans=1;
while(y)
{
if(y%2) ans*=x, ans%=MOD;
x*=x, x%=MOD, y/=2;
}
return ans;
}
ll N, L;
int a[5000];
ll ncr[3005][3005];
int main()
{
int i, j, k;
cin>>N>>L;
for(i=0;i<=N;i++) for(j=0;j<=i;j++) ncr[i][j]=j==0||j==i?1:ncr[i-1][j-1]+ncr[i-1][j], ncr[i][j]%=MOD;
for(i=0;i<N;i++) scanf("%d", &a[i]);
int n=count(a, a+N, 1);
ll ans=0;
for(k=0;k<=n;k++)
{
ans+=modpow(-1, k)*modpow(N-k, L)%MOD*ncr[n][k];
ans%=MOD;
}
ans*=mul_inv(modpow(N, L));
ans%=MOD;
ans+=MOD;
ans%=MOD;
cout<<ans;
return 0;
}
//*/
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