/*
  Markov連鎖の分布の時間発展の計算
  Time-stamp: "2021-05-19 Wed 09:19 JST hig"
*/
#define _CRT_SECURE_NO_WARNINGS /* Visual C++ に必要なおまじない */
#include <stdio.h>

#define NS 20

int multiply_trans(double pn[], double p[], int d);
int measure_dist(double p[], int t, int m);

int main(){
  int t,tmax;
  int x;
  double p[NS];
  double pn[NS];
  int m=NS;

  
  scanf("%d", &tmax);
  printf("#T=%d\n", tmax);

  /* ここに初期条件書いてね */
  t=0;
  for(x=0;x<m;x++){
    p[x]=0.0;
  }
  p[9]=1.0; //line-hidden
 
  measure_dist(p,t,m);
  
  for(t=1;t<=tmax;t++){
    
    multiply_trans(pn,p,m);

    for(x=0;x<m;x++){
      p[x]=pn[x];
    }

    measure_dist(p,t,m);
  
  }

  return 0;
}


/* int multiply_trans(double *pn, double *p, int m) を書いてね */ 
/* p に M をかけて pn=M p を書き込む */


int multiply_trans(double pn[], double p[],int m){
  int x;
  int start=0; //line-hidden
  int end=m-1; //line-hidden

  /* 周期的で簡単な場合 */
  for(x=start;x<end;x++){
    pn[x]=p[x+1];
  }
  pn[m-1]=p[0];

//begin-hidden
  /* 吸収壁. 端各2行, 計4行を特別扱い */
  for(x=start+2;x<end-1;x++){
    pn[x]=0.1*p[x-1]+0.5*p[x]+0.4*p[x+1];
  }
  pn[start]=                  0+ 1.0*p[start]  +0.4*p[start+1];
  pn[start+1]= 0.0*p[start+1-1]+0.5*p[start+1]+0.4*p[start+1+1];
  pn[end-1]=   0.1*p[end-1-1]  +0.5*p[end-1]  +0.0*p[end-1+1];
  pn[end]=     0.1*p[end-1]    +1.0*p[end]    +0;
  return 1;
  
  /* 反射壁の一種 */  
  for(x=start+2;x<end-1;x++){
    pn[x]=0.25*p[x-1]+0.5*p[x]+0.25*p[x+1];
  }
  pn[start]=                  0 +0.5*p[start]  +0.25*p[start+1];
  pn[start+1]= 0.5*p[start+1-1]+0.5*p[start+1]+0.25*p[start+1+1];
  pn[end-1]=   0.25*p[end-1-1]  +0.5*p[end-1]  +0.5*p[end-1+1];
  pn[end]=     0.25*p[end-1]    +0.5*p[end]    +0;
  return 1;

  /* 周期境界条件 */
  for(x=start+1;x<end;x++){
    pn[x]=0.1*p[x-1]+0.5*p[x]+0.4*p[x+1];
  }
  pn[start]=0.1*p[end]+  0.5*p[start]+0.4*p[start+1];
  pn[end]=  0.1*p[end-1]+0.5*p[end]  +0.4*p[start];
  return 1;
//end-hidden
}


/* tとpを1行に出力*/
int measure_dist(double p[], int t, int m){
  int x;
  printf("%d,",t);
  for(x=0;x<m;x++){
    printf("%.14f, ",p[x]);
  }
  printf("\n");
  return 0;
}