HDU3652(数位dp)

A - B-number

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Description

A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.

Input

Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).

Output

Print each answer in a single line.

Sample Input

13
100
200
1000

Sample Output

1
1
2
2

数位dp

 //2016.8.7
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>

 using namespace std;

 int dp[][][][];
 //dp[i][j][k][p]表示方法数,i表示已处理的长度，j表示mod13的余数，k表示前缀是否相同(0为相同，1为不同),p为1表示前一个数为1,为2表示已出现13，否则为0
 int mi[];

 int cal(int x, int n, int p)
 {
     return x*mi[n]%p;
 }

 int judge(int x)//三态函数
 {
     if(x==)return ;
     else if(x==)return ;
     else return ;
 }

 int solve(string s)
 {
     memset(dp, , sizeof(dp));
     dp[][][][] = ;
     int len = s.size();
     for(int i = ; i < len; i++)
       for(int j = ; j < ; j++)
         for(int k = ; k < ; k++)
           for(int p = ; p < ; p++)
           {
               if(dp[i][j][k][p]!=)
               {
                   int l, r;
                   l = ; r = (k==?s[i]-'':);
                   for(int x = l; x <= r; x++)
                   {
                       if(p==)
                       dp[i+][(j+cal(x, len-i-, ))%][(k==&&x==r)?:][x==?:]+=dp[i][j][k][p];
                       else if(p==)
                       dp[i+][(j+cal(x, len-i-, ))%][(k==&&x==r)?:][judge(x)]+=dp[i][j][k][p];
                       else if(p==)
                       dp[i+][(j+cal(x, len-i-, ))%][(k==&&x==r)?:][]+=dp[i][j][k][p];
                   }
               }
           }
     return dp[len][][][] + dp[len][][][];
 }

 int main()
 {
     string n;
     while(cin>>n)
     {
         mi[] = ;
         for(int i = ; i < ; i++)
           mi[i] = mi[i-]*%;
         int ans;
         ans = solve(n);
         cout<<ans<<endl;
     }

     return ;
 }