CF 1677D(Tokitsukaze and Permutations-冒泡排序)

Posted 2022-05-18 nike0good

已知长度为n的排列，经过k次冒泡（每次把最大的数交换到最后）后，得到的新序列为$a_i$.
$vi={\sum }_{j=1}^{i-1}=[a_i<a_j]$
现在已知$v_i$的某些地方的值，不知道的记$v_i=-1$，
求合法原排列数。

考虑$v_i$和排列达成双射关系。
且1次冒泡会导致$v_i$序列整体左移，并减1（若为0则不减）。最后添1位0
也即是

for(int i=1;i<n;i++)
	v[i]=min(v[i+1]-1,0);
v[n]=0;

因此$v$序列对应的原排列数为$k!\ast ({\sum }_{i=1}^{n-k}[v[i]==0](k+1))\ast ({\sum }_{i=1}^{n-k}[v[i]==-1](k+i))$
无解条件为
$(\forall i\le n-k,v[i]>i-1)\vee (\forall i>n-k,v[i]>0)$

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F ( 998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n)  \\
						For(j,m-1) cout<<a[i][j]<<' ';\\
						cout<<a[i][m]<<endl; \\
						 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b)return (a*b)%F;
ll add(ll a,ll b)return (a+b)%F;
ll sub(ll a,ll b)return ((a-b)%F+F)%F;
void upd(ll &a,ll b)a=(a%F+b%F)%F;
inline int read()

	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) if (ch=='-') f=-1; ch=getchar();
	while(isdigit(ch))  x=x*10+ch-'0'; ch=getchar();
	return x*f;
 
int n,m;
int v[1123456];
#define MAXN (1123456)
ll inj[MAXN],jie[MAXN];
inline int C(int a,int b) 
	return (ll)jie[a]*inj[b]%F*inj[a-b]%F;
inline int P(int a,int b) 
	return (ll)jie[a]*inj[a-b]%F;

ll p=F;
inline int pow2(int a,ll b)  //a^b mod p 
  
    if (b==0) return 1%p;  
    int c=pow2(a,b/2);  
    c=(ll)c*c%p;  
    if (b&1) c=(ll)c*a%p;  
    return c;  
  
void pre(int n) 
	jie[0]=1;For(i,n) jie[i]=mul(jie[i-1],i);
	inj[0]=inj[1]=1;Fork(i,2,n) inj[i]=(F-(F/i))*inj[F%i]%F;
	For(i,n) inj[i]=mul(inj[i],inj[i-1]);  
 
ll a[1123456];
ll calc() 
	int n=read(),k=read();
	For(i,n) v[i]=read();
	Fork(i,n-k+1,n) if(v[i]>0) return 0;
	For(i,n-k) if(v[i]>i-1) return 0;
	ll p=jie[k];
	For(i,n-k) if(v[i]==0) p=p*(k+1)%F;
	else if(v[i]==-1) p=p*(i+k)%F;
	return p;


int main()

//	freopen("D.in","r",stdin);
//	freopen(".out","w",stdout);
	pre(1e6);
	int T=read();
	while(T--) 
		cout<<calc()<<endl;
	
	return 0;