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【2023——二项式反演】

news 来源：原创 2025/6/12 17:19:12

题目

分析

采用至少->恰好

勘误：n-3k是位置，包括自由位数和潜在的4位段的位置

$f(i) =\binom{n-3i}{i} \cdot 10^{n-4i}$

$g(m) = \sum_{i=m}^{n/4} \binom{i}{m} \cdot (-1)^{i-m} \cdot f(i)$

代码

#include <iostream>
using namespace std;
using ll = long long;

const int N = 1e5+10;
const int mod = 998244353;

int fac[N], infac[N];

int qmi(int base, int expo)
{
  int retv = 1;
  while(expo)
  {
    if(expo & 1) retv = 1ll * retv * base % mod;
    base = 1ll * base * base % mod;
    expo >>= 1;
  }

  return retv;
}
void init(int n)
{
  fac[0] = 1, infac[0] = 1;
  for(int i = 1; i <= n; i++)
    fac[i] = 1ll * fac[i - 1] * i % mod;
  infac[n] = qmi(fac[n], mod-2);
  for(int i = n-1; i; i--)
    infac[i] = 1ll * infac[i+1] * (i+1) % mod; //注意这里是 *(i+1)
}
int C(int a, int b)
{
  return 1ll * fac[a] * infac[b] % mod * infac[a-b] % mod;
}
int main()
{
  init(1e5);

  int n, m;
  cin >> n >> m;

  ll ans = 0;
  for(int i = m; i <= n/4; i++)
  {
    ll seg = 1ll * C(i, m) * C(n-3*i, i) % mod * qmi(10, n-4*i) % mod;
    if(i-m&1) seg = -seg;
    ans = (ans + seg + mod) % mod; //存在-seg，要防负
  }

  cout << ans;
  return 0;
}