题目链接

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题目大意：

就是你有一个 n n n的全排列，现在问你去重排这个排列使得对于给定的 k k k,满足对于任意的 a i ， i > k a_i，i>k ai​，i>k的都有 a i > m i n ( a i − 1 , . . . , a i − k ) a_i>min(a_{i-1},...,a_{i-k}) ai​>min(ai−1​,...,ai−k​)
求排列数

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解题思路：

这里很妙的一点就是我们知道对于最小的数肯定是一定要放到 [ 1 , k ] [1,k] [1,k]里面去的，然后最小的数肯定会把数列切成两半，前面那一半是随便选，后面剩下的数又变成新的问题。
那么就是新的问题
那么 d p [ i ] = ∑ j = 1 k A i − 1 j − 1 d p [ i − j ] dp[i]=\sum_{j=1}^{k} A^{j-1}_{i-1} dp[i-j] dp[i]=∑j=1k​Ai−1j−1​dp[i−j]
预处理组合数然后前缀和维护 d p dp dp值就行了

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AC code

#include <bits/stdc++.h>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define LLF 0x3f3f3f3f3f3f3f3f
#define f first
#define s second
#define endl '\n'
using namespace std;
const int N = 1e7 + 10, mod = 998244353;
const int maxn = 500010;
const long double eps = 1e-5;
const int EPS = 500 * 500;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x) {
   x = 0;char ch = getchar();ll f = 1;
   while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
   while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {
   read(first);
   read(args...);
}
int n, k;
inline ll qmi(ll a, ll b) {
    ll res = 1;
    while(b) {
      if(b & 1) res = (res * a) % mod;
      b >>= 1;
      a = a * a % mod;
    }
    return res;
}

ll fac[N], inv[N], sum[N], dp[N];

inline void init() {
  fac[0] = 1;
  for(int i = 1; i < N; ++ i) fac[i] = (fac[i-1] * i) % mod;
  inv[N - 1] = qmi(fac[N-1],mod-2);
  for(int i = N - 2; i >= 0; -- i) inv[i] = inv[i+1] * (i+1) % mod;
}

int main() {
    IOS;
    init();
    cin >> n >> k;
    dp[0] = 1;
    sum[0] = 1;
    for(int i = 1; i <= n; ++ i) {
      dp[i] = (sum[i-1] - (i-k-1>=0?sum[i-k-1]:0) % mod + mod) % mod;
      dp[i] = dp[i] * fac[i-1] % mod;
      sum[i] = (sum[i-1] + (dp[i] * inv[i]) % mod) % mod;
    }
    cout << dp[n] % mod;
    return 0;
}