Codeforces Round #695 (Div. 2)

传送门

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D. Sum of Paths

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题目链接
中文翻译

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思路

\(f[i][j]\) 表示走了 \(j\) 步到 \(i\) 的所有方案数
那么 \(f[i][j] = f[i - 1][j - 1] + f[i + 1][j - 1]\)
由于路径可逆，\(f[i][j]\) 也可以表示从 \(i\) 走了 \(j\) 步的方案数

我们要求经过 \(i\) 的所有次数
\(times[i] = f[i][j] + f[i][k - j]\)
感性理解就是先从某处走了 \(j\) 步到 \(i\)，再从 \(i\) 走 \(k-j\) 步

那么此题就结束了 qwq

代码如下

点击查看代码

/********************
Author:  Nanfeng1997
Contest: Codeforces Round #695 (Div. 2)
URL:     https://codeforces.com/contest/{getProblemIndexes(problemCurrentPageList[i][0])[0]}/problem/D
When:    2022-03-15 20:45:20

Memory:  1024MB
Time:    3000ms
********************/
#include  <bits/stdc++.h>
using namespace std;
using LL = long long;
const int MOD = 1e9 + 7;
void solve() {

  int n, k, q; cin >> n >> k >> q;
  vector<int> w(n + 1); for(int i = 1; i <= n; i ++ ) cin >> w[i];
  
  vector<vector<LL> > f(n + 2, vector<LL> (k + 1, 0)); for(int i = 1; i <= n; i ++ ) f[i][0] = 1;
  for(int j = 1; j <= k; j ++ ) for(int i = 1; i <= n; i ++ ) {
      f[i][j] = (f[i - 1][j - 1] + f[i + 1][j - 1]) % MOD;
    }

  vector<LL> times(n + 1);

  for(int i = 1; i <= n; i ++ ) {
    for(int j = 0; j <= k; j ++ ) {
      times[i] = ((f[i][j] * f[i][k - j]) % MOD + times[i]) % MOD;
    }
  }

  LL ans = 0;
  for(int i = 1; i <= n; i ++ ) {
    ans = (ans + times[i] * w[i]) % MOD;
  }

  while(q -- ) {
    int id, x; cin >> id >> x; 
    ans = (ans - (times[id] * w[id]) % MOD + MOD) % MOD;
    ans = (ans + (times[id] * x) % MOD) % MOD;
    w[id] = x;
    cout << ans << "\n";
  }


}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();

  return 0;
}