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题解

看起来是一个很显然的矩阵乘法，然后看起来很显然的时间复杂度过大

但是写出来之后，发现这是一个循环矩阵

两个循环矩阵乘起来还是循环矩阵，所以只需要$O(n^2)$计算第一行就可以了

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
#define mod 1000000007ll
struct Matrix
{
	ll n, m, a[210][210];
	ll* operator[](ll x){return a[x];}
	Matrix(ll x, ll y)
	{
		n=x, m=y;
        ll i, j;
		rep(i,1,n)rep(j,1,m)a[i][j]=0;
	}
    Matrix(vector<vector<ll>> v)
    {
        ll i, j;
        n=v.size(), m=v[0].size();
        rep(i,1,n)rep(j,1,m)a[i][j]=v[i-1][j-1]%mod;
    }
    friend Matrix operator*(Matrix a, Matrix b)
    {
        Matrix t(a.n,b.m);
        ll i, j, k;
        rep(i,1,t.n)rep(k,1,a.m)rep(j,1,t.m)(t[i][j]+=a[i][k]*b[k][j])%=mod;
        return t;
    }
	void show()
	{
		ll i, j;
        cerr<<"Matrix "<<n<<"*"<<m<<":"<<endl;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)cerr<<a[i][j];
			cerr<<endl;
		}
	}
};
Matrix mult(Matrix a, Matrix b)
{
    ll n = a.n, i, j, k;
    Matrix c(n,n);
    rep(j,1,n)rep(k,1,n)
    {
        (c[1][j] += a[1][k]*b[j][k]) %=mod;
    }
    rep(i,2,n)
    {
        c[i][1] = c[i-1][n];
        rep(j,2,n)c[i][j] = c[i-1][j-1];
    }
    return c;
}
Matrix pow(Matrix a, ll b)
{
    Matrix ans(a.n,a.n), t=a;
    ll i, n=a.n;
    rep(i,1,n)ans[i][i]=1;
    for(;b;b>>=1,t=mult(t,t))if(b&1)ans=mult(ans,t);
    return ans;
}
int main()
{
    ll T=read();
    while(T--)
    {
        ll n=read(), m=read(), k=read(), i, j;
        Matrix a(n,1), A(n,n);
        rep(i,1,n)a[i][1]=read();
        rep(i,1,n)rep(j,1,n)
            if(i!=j and min(abs(i-j),n-abs(i-j))<k)A[i][j]=k-min(abs(i-j),n-abs(i-j));
        A = pow(A,m);
        // A.show();
        auto f = A * a;
        rep(i,1,n)
            printf("%lld ",f[i][1]);
        putchar(10);
    }
    return 0;
}