题面

题解

这数据范围……这输出大小……这模数……太有迷惑性了……

首先对于\(0\)来说，不管怎么选它们的排名都不会变，这个先特判掉

对于一个\(a_i\)来说，如果它不选，那么所有大于等于它的数随便选，乘\(2\)之后还是小于它的数也随便选

如果它选呢？所有大于等于它，且小于它的\(2\)倍的数全都得选，剩下的数就随便选不选了

然后没有然后了

//minamoto#include
    
     #define R register#define fp(i,a,b) for(R int i=a,I=b+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}char sr[1<<21],z[20];int K=-1,Z=0;inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}void print(R int x){    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++K]=z[Z],--Z);sr[++K]='\n';}const int N=1e5+5,P=998244353;inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}int ksm(R int x,R int y){    R int res=1;    for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);    return res;}int ans[N],a[N],b[N],sum[N],fac[N],ifac[N];int n,k,m,g;inline int C(R int n,R int m){return (m>n||m<0)?0:1ll*fac[n]*ifac[m]%P*ifac[n-m]%P;}int main(){//  freopen("testdata.in","r",stdin);    n=read(),k=read();    fac[0]=ifac[0]=1;    fp(i,1,n)fac[i]=mul(fac[i-1],i);    ifac[n]=ksm(fac[n],P-2);fd(i,n-1,1)ifac[i]=mul(ifac[i+1],i+1);    fp(i,1,n)a[i]=b[i]=read();    sort(b+1,b+1+n),m=unique(b+1,b+1+n)-b-1;    fp(i,1,n)a[i]=lower_bound(b+1,b+1+m,a[i])-b,++sum[a[i]];    fp(i,1,m)sum[i]+=sum[i-1];    b[1]==0?ans[1]=C(n,k):0;    for(R int i=(b[1]==0)?2:1,j=0,l=0;i<=m;++i){        while(j
      
       <<1)