bzoj 5416 noi 2018 冒泡排序

 1 #include 2 using namespace std; 
 3 int const N=1000000+10;  
 4 int const mod=998244353;  
 5 int n,a[N],pw[N1],inv[N1],vis[N];       
 6 int ksm(int x,int y){
 7     int ret=1;  
 8     while (y){
 9         if(y&1) ret=1LL*ret*x%mod;  
10         x=1LL*x*x%mod;  
11         y>>=1;
12     }
13     return ret;  
14 }
15 int calc(int x,int y){
16     if(xreturn 0;  
17     return 1LL*pw[x]*inv[y]%mod * inv[x-y]%mod;  
18 }
19 int main(){
20     //freopen("inverse.in","r",stdin);  
21     //freopen("inverse.out","w",stdout);  
22     int cas;  
23     scanf("%d",&cas);
24     pw[0]=1;  
25     for(int i=1;i2000000;i++)  
26         pw[i]=1LL*pw[i-1]*i%mod;  
27     inv[2000000]=ksm(pw[2000000],mod-2);  
28     for(int i=1999999;i>=0;i--)  
29         inv[i]=1LL*inv[i+1]*(i+1)%mod;    
30     while (cas--){
31         scanf("%d",&n);  
32         for(int i=1;i)  
33             scanf("%d",&a[i]);
34         int mv=0,ans=0,mn=1;   
35         memset(vis,0,sizeof(vis));   
36         for(int i=1;i){ 
37             int lim=max(mv+1,a[i]+1); 
38             vis[a[i]]=1;   
39             if(limn)  
40                 ans=((ans+calc(2*n-i-lim+1,n-i+1)-calc(2*n-i-lim+1,n-i+2))%mod + mod) % mod; 
41             if(a[i]>mv) mv=a[i];  
42             else if(a[i]!=mn)  break;  
43             while (vis[mn]) mn++;     
44         }  
45         printf("%d\n",ans);  
46 
47     }
48     return 0; 
49 }