hdu3652B-number

Problem 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

 

Author

wqb0039

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

题解：

设f[i][j][r][0/1]表示i位数（可含前导0）第一位为j，mod13的余数是r，有没有出现13(0/1)的数有几个

code：

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int maxl=;
 const int maxnum=;
 const int rest=;
 int n,power[maxl],f[maxl][maxnum][rest][],a[maxl];
 void init(){
     power[]=;
     for (int i=;i<=;i++) power[i]=power[i-]*%;
     f[][][][]=;
     for (int i=;i<=;i++) for (int j=;j<=;j++)
         for (int k=;k<=;k++) for (int r=;r<;r++){
             int tmp=(r+j*power[i])%;
             if (!(j==&&k==)) f[i][j][tmp][]+=f[i-][k][r][];
             else f[i][j][tmp][]+=f[i-][k][r][];
             f[i][j][tmp][]+=f[i-][k][r][];
         }
 }
 int calc(int n){
     int ans=,t=n,len=,r=,tmp;
     bool flag=;
     memset(a,,sizeof(a));
     while (t) a[++len]=t%,t/=;
     for (int i=len;i;i--){
         for (int j=;j<a[i];j++){
             tmp=((-(r+j*power[i]))%+)%;
             for (int k=;k<=;k++) ans+=f[i-][k][tmp][];
             if (!flag){
                 if (a[i+]==){if (j==) for (int k=;k<=;k++) ans+=f[i-][k][tmp][];}
                 if (j==) ans+=f[i-][][tmp][];
             }
             else for (int k=;k<=;k++) ans+=f[i-][k][tmp][];
         }
         r+=a[i]*power[i],r%=;
         if (a[i+]==&&a[i]==) flag=;
     }
     return ans;
 }
 int main(){
     init();
     while (~scanf("%d",&n)) printf("%d\n",calc(n+));
     return ;
 }