Problem:
Score : 500 points

Problem Statement

Find the number of integers between 1 and N (inclusive) that contains exactly K non-zero digits when written in base ten.

Constraints

1 \leq N < 10^{100}

1 \leq K \leq 3

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the count.

Sample Input 1

100
1

Sample Output 1

19

The following 19 integers satisfy the condition:

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100

Sample Input 2

25
2

Sample Output 2

14

The following 14 integers satisfy the condition:

11,12,13,14,15,16,17,18,19,21,22,23,24,25

Sample Input 3

314159
2

Sample Output 3

937

Sample Input 4

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
3

Sample Output 4

117879300

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
#include<assert.h>
#include<ctype.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
#define INF (1LL<<60)
#define MOD1 1000000007
#define MOD2 998244353
#define MAX_N (1 << 20)
#define YES printf("Yes\n")
#define NO printf("No\n")
#define PN printf("\n")
#define charsize 100005 //10^5+5
#define PI 3.141592653589793238

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a= c;}
void cin(ll *n){ scanf("%lld",&(*n)); }
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll min3(ll a, ll b, ll c){return (a<=b && a<=c) ? a : b<=c ? b : c;}
ll max3(ll a, ll b, ll c){return (a>=b && a>=c) ? a : b>=c ? b : c;}
ll minn(ll n, ll a[]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
double POW_d(double a, double b){double c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod_MOD1(ll n){n+= n<0?((-n)/MOD1+1)*MOD1:0; return n%=MOD1;}
ll mod_p(ll n ,ll p){n+= n<0?((-n)/p+1)*p:0; return n%=p;}
ll change_into_num(char s[] , ll len, ll p){ return !p ? 0 : POW(10,p-1)*(s[len-p]-'0') + change_into_num(s,len,p-1); }
ll digits(ll a, ll b){return a/b?1+digits(a/b,b):1;}
ll base(ll n, ll a, ll i){return i==1?n%a:base(n/a,a,i-1);}

void lr_lower( int *l, int *r, ll am, ll val , int type ){ (type<3) ? ( am < val ?  ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ) : ( am <= val ? ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ); }
void lr_upper( int *l, int *r, ll am, ll val , int type ){ (type<3) ? ( am <= val ?  ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ) : ( am < val ? ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ); }
int cmp_lower( ll a, ll b, int type ){ return (type==1) ? ( a==b ? 1 : 0 ) : (type==2) ? ( a>=b ? 1 : 0 ) : ( a>b ? 1 : 0 ) ; }
int cmp_upper( ll a, ll b, int type ){ return (type==1) ? ( a==b ? 1 : 0 ) : (type==2) ? ( a<=b ? 1 : 0 ) : ( a<b ? 1 : 0 ) ; }
// return smallest p  which meets  a[p]==val :1  >=:2   >:3
ll lower_bound( ll a[], int l, int r, ll val , int type ){  while(r-l>1) lr_lower(&l,&r,a[ (l+r)/2 ],val,type);  return cmp_lower(a[l],val,type) ? l : cmp_lower(a[r],val,type) ? r : -1;  }
// return biggest p  which meets   a[p]==val :1  <=:2   <:3
ll upper_bound( ll a[], int l, int r, ll val , int type ){  while(r-l>1) lr_upper(&l,&r,a[ (l+r)/2 ],val,type); return cmp_upper(a[r],val,type) ? r : cmp_upper(a[l],val,type) ? l : -1; }
// count i  which meets ai==x
ll count(ll a[], int l, int r, ll x){  int p = lower_bound(a,l,r,x,1);  return p==-1 ? 0 : upper_bound(a,p,r,x,1)-p+1; }

ll *factors[2] , fac_cnt=0 , is_factor_prepared=0;
ll factor_pre(){  rep(i,0,1){ if(is_factor_prepared++) return 0; }  ll tmp=(1e6)/2+1, fac_tmp[tmp];  rep(i,0,tmp){fac_tmp[i]=i?2*i+1:2;}  rep(i,1,tmp){if(fac_tmp[i]){  repp(j,3,tmp/(2*i+1)+1,2 ){  if( j*(2*i+1)<tmp ) fac_tmp[ (j*(2*i+1)-1)/2 ]=0;  }  }else continue;}  rep(i,0,tmp){if(fac_tmp[i]){  rep(j,0,2){ factors[j] = realloc( factors[j] , sizeof(ll)*( fac_cnt +1 ) );  factors[j][j?fac_cnt++:fac_cnt]=j?0:fac_tmp[i];  }  } }  return 0;  }
ll factor(ll n, ll new_common_plus){ factor_pre();  rep(i,0,fac_cnt){ ll cnt=0; rep(j,0,1){ while( ( cnt+= n %factors[0][i]==0 ? 1 : 0 )  && (n/=factors[0][i]) %factors[0][i]==0 ) continue; } factors[1][i]= new_common_plus==1 ? cnt :  new_common_plus==2 ? max2(factors[1][i],cnt) : factors[1][i]+cnt ;  if( factors[0][i]> n ) break; } return n; }
ll judge_prime(ll n){ factor_pre(); rep(i,0,fac_cnt){ if(n<factors[0][i]*factors[0][i] || n==factors[0][i]) break; else if(n%factors[0][i]==0) n/=n; }  return n==1?0:1; }

ll *mf_arr,*inv_arr,*finv_arr,is_minv_made=0,is_mf_made=0;
ll makeinv(ll n , ll mod){  rep(i,0,1){if(is_minv_made++) return 0;}  inv_arr = realloc(inv_arr, sizeof(ll)*2 );  finv_arr = realloc(finv_arr, sizeof(ll)*2 );  inv_arr[1]=1;finv_arr[0]=finv_arr[1]=1;  rep(i,2,n+1){  inv_arr = realloc(inv_arr, sizeof(ll)*(i+1) );  finv_arr = realloc(finv_arr, sizeof(ll)*(i+1) );  inv_arr[i]= mod - inv_arr[mod%i] * (mod / i) % mod;  finv_arr[i] = finv_arr[i - 1] * inv_arr[i] % mod;  }  return 0; }
ll make_mf(ll n, ll mod){  rep(i,0,1){ if(is_mf_made++) return 0; }  mf_arr = realloc(mf_arr, sizeof(ll)*2 );  ll x=1;  mf_arr[0]=mf_arr[1]=x;  rep(i,2,n+1){   x=x*i%mod;  mf_arr = realloc(mf_arr, sizeof(ll)*(i+1) );  mf_arr[i]=x; }  return 0;  }
ll m_inv(ll x, ll mod, ll is_fac ){ makeinv(2*1e6+10,mod);  return is_fac?finv_arr[x]:inv_arr[x]; }
ll m_f(ll x, ll mod){  make_mf(2*1e6+10,mod);  return mf_arr[x];  }
// ll mod_nck(ll n, ll k, ll mod){ return m_f[n]*finv_arr[k]%mod*finv_arr[n-k]%mod; }
ll m_p(ll r,ll n,ll mod){  ll t=1,s=r;  while(n>0){ t = (n&1) ? t*s%mod : t;  s=s*s%mod; n>>=1; }  return r?t:0;  }
ll m_mul2(ll a, ll b, ll mod){ return a*b%mod; }
ll m_mul3(ll a, ll b, ll c, ll mod){ return m_mul2(a*b%mod,c,mod); }
ll m_mul4(ll a, ll b, ll c, ll d, ll mod){ return m_mul3(a*b%mod,c,d,mod); }
ll m_mul5(ll a, ll b, ll c, ll d, ll e, ll mod){ return m_mul4(a*b%mod,c,d,e,mod); }

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
int cmp_string( const void * a , const void * b ) {  return strcmp( (char *)a , (char *)b ); }  // qsort((void*)s,n,sizeof(s[0]),int_sort );
int cmp_char(const void * a, const void * b) { return *(char *)a - *(char *)b;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
void sort_string(int n,int size,char s[][size]){ qsort( (void*)s , n , sizeof(s[0]) , cmp_string ); }
void sort_char(char *s){ qsort( (void *)s , strlen(s) , sizeof(char) , cmp_char ); }
ll unique_string(ll n ,ll size, char s[][size]){ ll ans=1; rep(i,1,n) if( strcmp(s[i],s[i-1]) ) ans++; return ans; }
ll unique_num(ll n , ll a[]){ ll ans=1; rep(i,1,n) if( a[i]!=a[i-1] ) ans++; return ans; }

typedef struct{ ll a , b;}fr;
int cmp1( const void *p, const void *q ) { return ((fr*)p) ->a - ((fr*)q)->a;}
int cmp2( const void *p, const void *q ) { return ((fr*)q) ->a - ((fr*)p)->a;}
void strsortup(fr*a,int n){qsort(a,n,sizeof(fr),cmp1);}
void strsortdown(fr*a,int n){qsort(a,n,sizeof(fr),cmp2);}




char s[1151154];
ll dp[105][2][5]={};
int main(void){
    // fgets(s,sizeof(s),stdin); 
    ll n,k;
    // ll ans=0;
    scanf("%s",s);
    // cin(&n);
    n=strlen(s);
    cin(&k);
    dp[0][1][0]=1;
    rep(i,0,n){
        ll p=s[i]-'0';
        rep(k,0,4){
            rep(j,0,10){
                if(j<p){
                    if(!j){
                        dp[i+1][0][k]+=dp[i][0][k]+dp[i][1][k];
                    }else{
                        dp[i+1][0][k+1]+=dp[i][0][k]+dp[i][1][k];
                    }
                }
                if(j==p){
                    if(!j){
                        dp[i+1][1][k]+=dp[i][1][k];
                        dp[i+1][0][k]+=dp[i][0][k];
                    }else{
                        dp[i+1][1][k+1]+=dp[i][1][k];
                        dp[i+1][0][k+1]+=dp[i][0][k];
                    }
                }
                if(j>p){
                    dp[i+1][0][k+1]+=dp[i][0][k];
                }
            }
        }
    }

    printf("%lld\n",dp[n][1][k]+dp[n][0][k]);
    return 0;
}