牛客 109 C 操作数 (组合数学)

给定长度为n的数组a，定义一次操作为：
1. 算出长度为n的数组s，使得s_i= (a[1] + a[2] + ... + a[i]) mod 1,000,000,007；
2. 执行a = s；
现在问k次操作以后a长什么样

记初始序列为$a_0$, $k$次操作后序列为$a_k$, 有

$egin{equation}
a_k
=egin{bmatrix}
1 & 0 & 0 &cdots &0 &0\
1 & 1 & 0 &cdots &0 &0\
1 & 1 & 1 & cdots & 0 & 0\
vdots & vdots & vdots & ddots & vdots &vdots\
1 & 1 & 1 &cdots &1& 1\
end{bmatrix}^k a_0
end{equation}$

矩阵幂可以化简为

$egin{equation}
egin{bmatrix}
inom{k-1}{0} & 0 & 0 &cdots &0 &0\
inom{k}{1} & inom{k-1}{0} & 0 &cdots &0 &0\
inom{k+1}{2} & inom{k}{1} &inom{k-1}{0} & cdots & 0 & 0\
vdots & vdots & vdots & ddots & vdots &vdots\
inom{k+n-2}{n-1} & inom{k+n-3}{n-2} & inom{k+n-4}{n-3} &cdots &inom{k}{1}& inom{k-1}{0}\
end{bmatrix}
end{equation}$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 2e3+10;
int n, k;
int a[N], f[N];

int main() {
	scanf("%d%d", &n, &k);
	REP(i,1,n) scanf("%d", a+i);
	f[1] = 1;
	REP(i,2,n) f[i]=(ll)f[i-1]*(k+i-2)%P*inv(i-1)%P;
	REP(i,1,n) {
		int ans = 0;
		REP(j,1,i) ans = (ans+(ll)f[i-j+1]*a[j])%P;
		printf("%d ", ans);
	} hr;
}