【计数】cf223C. Partial Sums

考试时候遇到这种题只会找规律

You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that:

  1. First we build by the array a an array s of partial sums, consisting of n elements. Element number i (1 ≤ i ≤ n) of array s equals . The operation x mod y means that we take the remainder of the division of number x by number y.
  2. Then we write the contents of the array s to the array a. Element number i (1 ≤ i ≤ n) of the array s becomes the i-th element of the array a (ai = si).

You task is to find array a after exactly k described operations are applied.

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 109). The next line contains n space-separated integers a1, a2, ..., an — elements of the array a (0 ≤ ai ≤ 109).

Output

Print n integers  — elements of the array a after the operations are applied to it. Print the elements in the order of increasing of their indexes in the array a. Separate the printed numbers by spaces.


题目分析

可以从矩阵乘法开始想起,考虑转移矩阵,发现其主对角线下方全为0、元素按照次对角线对称。

或者就是找规律

3.16upd:

总觉得一个经典模型不应该用找规律这么假的方式随随便便搞掉吧……

网上的题解都是打表找规律 | 矩乘找规律 | dp找规律……

来自ZZK的新的理解方式:$a_i$的贡献也就是它走到$s^p_j$的方案数量。

 1 #include<bits/stdc++.h>
 2 #define MO 1000000007
 3 const int maxn = 2035;
 4 
 5 int n,k,a[maxn],b[maxn],inv[maxn],fac[maxn],pre[maxn];
 6 
 7 void init()
 8 {
 9     fac[0] = fac[1] = inv[0] = inv[1] = 1;
10     for (int i=2; i<=n; i++)
11         fac[i] = 1ll*fac[i-1]*i%MO, 
12         inv[i] = MO-1ll*(MO/i)*inv[MO%i]%MO;
13     pre[0] = 1;
14     for (int i=1; i<=n; i++)
15         pre[i] = 1ll*pre[i-1]*(k%MO+i-1)%MO*inv[i]%MO;
16 }
17 int main()
18 {
19     scanf("%d%d",&n,&k);
20     for (int i=1; i<=n; i++) scanf("%d",&a[i]), b[i] = a[i];
21     if (k){
22         init();
23         for (int i=1; i<=n; i++)
24         {
25             b[i] = 0;
26             for (int j=1; j<=i; j++)
27                 b[i] = 1ll*(b[i]+1ll*pre[i-j]*a[j]%MO)%MO;
28         }
29     }
30     for (int i=1; i<=n; i++) printf("%d ",b[i]);
31     return 0;
32 }

 

 

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posted @ 2019-03-15 21:25  AntiQuality  阅读(147)  评论(0编辑  收藏