1 // 多校5 HDU5787 K-wolf Number 数位DP
2 // dp[pos][a][b][c][d][f] 当前在pos，前四个数分别是a b c d
3 // f 用作标记，当现在枚举的数小于之前的数时，就不用判断i与dig[pos]的大小
4 // 整体来说就，按位往后移动，每次添加后形成的数都小于之前的数，并且相邻k位不一样，一直深搜到cnt位
5 // http://blog.csdn.net/weizhuwyzc000/article/details/52097690
6
7
8
9 // #pragma comment(linker, "/STACK:102c000000,102c000000")
10 #include <iostream>
11 #include <cstdio>
12 #include <cstring>
13 #include <sstream>
14 #include <string>
15 #include <algorithm>
16 #include <list>
17 #include <map>
18 #include <vector>
19 #include <queue>
20 #include <stack>
21 #include <cmath>
22 #include <cstdlib>
23 // #include <conio.h>
24 using namespace std;
25 #define clc(a,b) memset(a,b,sizeof(a))
26 const double inf = 0x3f3f3f3f;
27 #define lson l,mid,rt<<1
28 // #define rson mid+1,r,rt<<1|1
29 const int N = 2010;
30 const int M = 1e6+10;
31 const int MOD = 1e9+7;
32 #define LL long long
33 #define LB long double
34 // #define mi() (l+r)>>1
35 double const pi = acos(-1);
36 const double eps = 1e-8;
37 void fre(){freopen("in.txt","r",stdin);}
38 void freout(){freopen("out.txt","w",stdout);}
39 inline int read(){int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;}
40
41 LL l,r,cnt;
42 int k;
43 int dig[19];
44 LL dp[19][11][11][11][11][2];
45 int vis[19][11][11][11][11][2];
46 int cas;
47 LL dfs(LL pos,int a,int b,int c,int d,int f){
48 if(pos==cnt) return 1LL;
49 if(vis[pos][a][b][c][d][f]==cas) return dp[pos][a][b][c][d][f];
50 vis[pos][a][b][c][d][f]=cas;
51 LL ans=0;
52 if(f){
53 for(int i=0;i<10;i++){
54 if(k==2) {if(d==i) continue;}
55 else if(k==3) {if(d==i||c==i) continue;}
56 else if(k==4) {if(d==i||c==i||b==i) continue;}
57 else {if(d==i||c==i||b==i||a==i) continue;}
58 if(i==0){
59 if(d==10) ans+=dfs(pos+1,a,b,c,d,f);
60 else ans+=dfs(pos+1,b,c,d,i,f);
61 }
62 else ans+=dfs(pos+1,b,c,d,i,f);
63 }
64 }
65 else{
66 for(int i=0;i<10;i++){
67 if(k==2) {if(d==i) continue;}
68 else if(k==3) {if(d==i||c==i) continue;}
69 else if(k==4) {if(d==i||c==i||b==i) continue;}
70 else {if(d==i||c==i||b==i||a==i) continue;}
71 if(i<dig[pos]){
72 if(i==0){
73 if(d==10) ans+=dfs(pos+1,a,b,c,d,1);
74 else ans+=dfs(pos+1,b,c,d,i,1);
75 }
76 else ans+=dfs(pos+1,b,c,d,i,1);
77 }
78 else if(i==dig[pos]){
79 if(i==0){
80 if(d==10) ans+=dfs(pos+1,a,b,c,d,f);
81 else ans+=dfs(pos+1,b,c,d,i,f);
82 }
83 else
84 ans+=dfs(pos+1,b,c,d,i,f);
85 }
86 }
87 }
88 return dp[pos][a][b][c][d][f]=ans;
89 }
90
91 bool ck(){
92 for(int i=0;i<cnt;i++){
93 for(int j=i-1;j>=max(0,i-k+1);j--){
94 if(dig[i]==dig[j]) return false;
95 }
96 }
97 return true;
98 }
99 int tem[19];
100 int main(){
101 while(~scanf("%I64d%I64d%d",&l,&r,&k)){
102 cnt=0;
103 while(r) tem[cnt++]=r%10,r=r/10;
104 int k=0;
105 for(int i=cnt-1;i>=0;i--) dig[k++]=tem[i];
106 cas++;
107 LL ans1=dfs(0,10,10,10,10,0);
108 cnt=0,k=0;
109 while(l) tem[cnt++]=l%10,l=l/10;
110
111 for(int i=cnt-1;i>=0;i--) dig[k++]=tem[i];
112 cas++;
113 LL ans2=dfs(0,10,10,10,10,0);
114 if(ck()) ans2--;
115 printf("%I64d\n",ans1-ans2);
116 }
117 return 0;
118 }
