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[12][13][2][3];
//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[12];

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

int judge(int x)//三态函数
{
    if(x==1)return 1;
    else if(x==3)return 2;
    else return 0;
}

int solve(string s)
{
    memset(dp, 0, sizeof(dp));
    dp[0][0][1][0] = 1;
    int len = s.size();
    for(int i = 0; i < len; i++)
      for(int j = 0; j < 13; j++)
        for(int k = 0; k < 2; k++)
          for(int p = 0; p < 3; p++)
          {
              if(dp[i][j][k][p]!=0)
              {
                  int l, r;
                  l = 0; r = (k==1?s[i]-'0':9);
                  for(int x = l; x <= r; x++)
                  {
                      if(p==0)
                      dp[i+1][(j+cal(x, len-i-1, 13))%13][(k==1&&x==r)?1:0][x==1?1:0]+=dp[i][j][k][p];
                      else if(p==1)
                      dp[i+1][(j+cal(x, len-i-1, 13))%13][(k==1&&x==r)?1:0][judge(x)]+=dp[i][j][k][p];
                      else if(p==2)
                      dp[i+1][(j+cal(x, len-i-1, 13))%13][(k==1&&x==r)?1:0][2]+=dp[i][j][k][p];
                  }
              }
          }
    return dp[len][0][0][2] + dp[len][0][1][2];
}

int main()
{
    string n; 
    while(cin>>n)
    {
        mi[0] = 1;
        for(int i = 1; i < 12; i++)
          mi[i] = mi[i-1]*10%13;
        int ans;
        ans = solve(n);
        cout<<ans<<endl;
    }

    return 0;
}