1 #include<stdio.h> 2 #include<string.h> 3 using namespace std; 4 typedef long long ll; 5 6 int a[20]; 7 int dp[20][15][3]; 8 9 //pos:当前位置 10 //mod：余数 11 //have ：0：前一位不是1 1：前一位是1 2：有13 12 ll dfs(int pos,int mod,int have,bool limit){ 13 if(pos==-1) return mod == 0 && have == 2; 14 if(!limit && dp[pos][mod][have] != -1) 15 return dp[pos][mod][have];//没有上限并且已被访问过 16 int up = limit?a[pos]:9; 17 ll ans = 0,i; 18 for(i = 0;i <= up;i++){ 19 int modx = (mod*10+i)%13; 20 int havex = have; 21 if(have == 0 && i==1) havex=1; 22 if(have == 1 && i!=1) havex=0; 23 if(have == 1 && i == 3) havex=2; 24 ans += dfs(pos-1, modx, havex, limit && i == a[pos]); 25 } 26 if(!limit) dp[pos][mod][have] = ans; 27 return ans; 28 } 29 30 ll solve(ll x){ 31 int pos=0; 32 while(x){ 33 a[pos++]=x%10; 34 x/=10; 35 } 36 return dfs(pos-1,0,0,true); 37 } 38 int main(){ 39 ll n; 40 memset(dp,-1,sizeof(dp)); 41 while(scanf("%lld",&n)!=EOF){ 42 ll ans = solve(n); 43 printf("%lld ",ans; 44 } 45 return 0; 46 }