題目鏈接

題意 : 中文題、點鏈接

分析 :

有道題是問你不斷求前綴和後的結果 Click here

這道題問的是逆過程

分析方法雷同、可參考 Click here

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正著做的矩陣是一個下三角

1 0 0 0

1 1 0 0

1 1 1 0

1 1 1 1

結合楊輝三角可得

C(k, 0)

C(k+1, 1) C(k, 0)

C(k+2, 2) C(k+1, 1) C(k, 0)

C(k+3, 3) C(k+2, 2) C(k+1, 1) C(k, 0)

......

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逆過程是這樣一個矩陣

1 0 0 0

-1 1 0 0

0 -1 1 0

0 0 -1 1

結合楊輝三角可得

A[i][j] = (-1)^(i-j) * C(k, i-j)

#include<bits/stdc++.h>
#define LL long long
#define ULL unsigned long long

#define scl(i) scanf("%lld", &i)
#define
 
 scll(i, j) scanf("%lld %lld", &i, &j)
#define sclll(i, j, k) scanf("%lld %lld %lld", &i, &j, &k)
#define scllll(i, j, k, l) scanf("%lld %lld %lld %lld", &i, &j, &k, &l)

#define scs(i) scanf("%s", i)
#define sci(i) scanf("%d", &i)
#define scd(i) scanf("%lf", &i)
#define
 
 scIl(i) scanf("%I64d", &i)
#define scii(i, j) scanf("%d %d", &i, &j)
#define scdd(i, j) scanf("%lf %lf", &i, &j)
#define scIll(i, j) scanf("%I64d %I64d", &i, &j)
#define sciii(i, j, k) scanf("%d %d %d", &i, &j, &k)
#define scddd(i, j, k) scanf("%lf %lf %lf", &i, &j, &k)
#define scIlll(i, j, k) scanf("%I64d %I64d %I64d", &i, &j, &k)
#define sciiii(i, j, k, l) scanf("%d %d %d %d", &i, &j, &k, &l)
#define scdddd(i, j, k, l) scanf("%lf %lf %lf %lf", &i, &j, &k, &l)
#define scIllll(i, j, k, l) scanf("%I64d %I64d %I64d %I64d", &i, &j, &k, &l)

#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define lowbit(i) (i & (-i))
#define mem(i, j) memset(i, j, sizeof(i))

#define fir first
#define sec second
#define VI vector<int>
#define ins(i) insert(i)
#define pb(i) push_back(i)
#define pii pair<int, int>
#define VL vector<long long>
#define mk(i, j) make_pair(i, j)
#define all(i) i.begin(), i.end()
#define pll pair<long long, long long>

#define _TIME 0
#define _INPUT 0
#define _OUTPUT 0
clock_t START, END;
void __stTIME();
void __enTIME();
void __IOPUT();
using namespace std;
const int maxn = 1e3 + 5;
const LL mod = 1e9 + 7;

LL arr[maxn];
LL A[maxn][maxn];
LL Comb[maxn];
LL inv[maxn];

inline void inv_init()
{
    inv[0] = inv[1] = 1;
    for(int i=2; i<maxn; i++)
        inv[i] = (LL)(mod - mod / i) * inv[mod % i] % mod;
}

int main(void){__stTIME();__IOPUT();

    inv_init();

    int n, k;
    int nCase;

    sci(nCase);
    while(nCase--){
        scii(n, k);

        LL tmp = 0;
        Comb[0] = 1LL;
        for(int i=1; i<=min(k, n); i++){
            Comb[i] = Comb[i-1]%mod;
            Comb[i] = ( Comb[i] * (k-i+1) ) % mod;
            Comb[i] = ( Comb[i] * inv[i] ) %mod;
        }

        for(int i=1; i<=n; i++) scl(arr[i]);
        for(int i=1; i<=n; i++){
            for(int j=1; j<=i; j++){
                if(i-j > k){
                    A[i][j] = 0LL;
                    continue;
                }

                if((i-j) & 1) A[i][j] = -1LL;
                else A[i][j] = 1LL;

                A[i][j] = ( ( (A[i][j] * Comb[i-j])%mod) + mod) % mod;
            }
        }

        for(int i=1; i<=n; i++){
            LL ans = 0;
            for(int j=1; j<=n; j++)
                ans = ((ans + (A[i][j] * arr[j])%mod + mod)%mod)%mod;
            printf("%lld", ans%mod);
            if(i != n) putchar(‘ ‘);
        }puts("");
    }















__enTIME();return 0;}


void __stTIME()
{
    #if _TIME
        START = clock();
    #endif
}

void __enTIME()
{
    #if _TIME
        END = clock();
        cerr<<"execute time = "<<(double)(END-START)/CLOCKS_PER_SEC<<endl;
    #endif
}

void __IOPUT()
{
    #if _INPUT
        freopen("in.txt", "r", stdin);
    #endif
    #if _OUTPUT
        freopen("out.txt", "w", stdout);
    #endif
}

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