Codeforces 223 C

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Combinatorics

Description

Given an array \(a[]\), define an operation:

$
s_i=\sum_{j=1}^{n}{a_j},\
s_i\rightarrow a_i.$

Calculate the value of each \(a\) after \(k\) times of operations.

Solution

The official tutorial wrote confusingly, so did other solutions on Internet.

Calculate how element \(a_j\) will affects \(a_i\) \((j \le i)\), after \(k\) times.

Let‘s draw something first.

\((i,j,k)\,\text{means}\, i \cdot a_1 + j \cdot a_2 + k \cdot a_3,\ \text{and}\,K \,\text{is times of operation}:\)

\(\begin{cases} (1,0,0)\, (1,1,0)\, (1,1,1)\quad \text{where K = 1,}\\ (1,0,0)\, (2,1,0)\, (3,2,1)\quad \text{where K = 2,}\\ (1,0,0)\, (3,1,0)\, (6,3,1)\quad \text{where K = 3},\\ \cdots \end{cases}\)

We also have another graph:

$
\begin{align}
1,1\
1,2,1\
1,3,3,1\
1,4,6,4,1\
\cdots
\end{align}
$
**cnblogs cannot start a new line in mathjax...

Think when \(k\) is changing, how the factor before \(a_j\) in \(a_i\) follows, we get:

\[fac_{i,j}^{k} = {j-i+k-1 \choose j-i}\]

To simplify our calculation, we found that \(fac_{i,j}^{k} = fac_{i-j+1,1}^{k}\), for we can replace \(j-i\) with \(j-i+1-1\), they are the same!

So our task now is to calculate \(fac_{i,1}^{k}\), which is \(i + k - 2 \choose i-1\). But we cannot pre process the factorial of \(k\), as \(k\) can be up to \(10^9\). But \(fac_{1,1}\) is always 1!

Think the transfer between \(fac_{i-1,1}\) and \(fac_{i,1}\), that is:

\[\frac{(i-1+k-2)!}{(i-1-1)!(k-1)!} \rightarrow \frac{(i+k-2)!}{(i-1)!(k-1)!}\]

We just need to multiply the first expression by \(i+k-2\) and \((i-1)^{-1}\,(mod\,10^9+7)\). And we can pre process all factors in \(O(n)\) time.

Then finish all calculation in \(O(n^2)\) time, as\(\begin{align}s_i=\sum_{j=1}^{i}{f_{i-j+1}\times a_j}\end{align}\).

Code

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
const ll p = 1e+9 + 7;
const int maxn = 2005;
int n, k;
ll inv[maxn], a[maxn], c[maxn], s[maxn];
void init() {
  inv[1] = 1;
  for (int i = 2; i <= 2000; ++i) {
    inv[i] = (-(p/i) + p) * inv[p % i] % p;
  }
}
int main() {
  init();
  ios::sync_with_stdio(false);
  cin.tie(0); cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];

  if(!k) {
    for(int i = 1; i <= n; ++i) s[i] = a[i];
  } else {
    c[1] = 1;
    for (int i = 2; i <= n; ++i) c[i] = c[i - 1] * (k + i - 2) % p * inv[i - 1] % p;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= i; ++j) s[i] = (s[i] + c[i - j + 1] * a[j] % p) % p;
  }
  for (int i = 1; i <= n; ++i) cout << s[i] << " ";
  cout << endl;
  return 0;
}

Codeforces 223 C

标签：pac   mat   ring   blog   internet   lse   multi   script   another   

原文地址：https://www.cnblogs.com/nishikino-curtis/p/9808312.html