hdu3652 (Digital dp)

The meaning of problems: seeking 1-n (n <= 1000000000) number, the skill is divisible by 13. The number 13 also includes a number of sub-strings;

Solution: Digital dp. LL dp [pre] [now] [have] [iflow] [rem] record of the pre bit now, have representation in front of whether there have been 13, iflow represents dp process whether it has to reduce, rem represents the remainder perhaps the need is.

Code:

/******************************************************
* author:xiefubao
*******************************************************/
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <string.h>
//freopen ("in.txt" , "r" , stdin);
using namespace std;

#define eps 1e-8
const double pi=acos(-1.0);
typedef long long LL;
const int Max=10100;
const int INF=1000000007;
LL dp[20][20][2][2][15];
int reminder[20];
int num[30];
LL dfs(int pre,int now,bool have,bool iflow,int rem)
{
    if(pre==0)
        return have&&rem==0;
    if(dp[pre][now][have][iflow][rem]!=-1)
        return dp[pre][now][have][iflow][rem];
    int en=iflow? 9 : num[pre-1];
    LL ans=0;
    for(int i=0;i<=en;i++)
    {
        int tool=((rem-(i*reminder[pre-1])%13)+13)%13;
        if(now==1&&i==3)
        ans+=dfs(pre-1,i,1,iflow||i!=en,tool);
        else
        ans+=dfs(pre-1,i,have,iflow||i!=en,tool);
    }
    return dp[pre][now][have][iflow][rem]=ans;
}
void init()
{
    reminder[0]=1;
    for(int i=1;i<=10;i++)
        reminder[i]=(reminder[i-1]*10)%13;
}
LL solve(LL n)
{
    int p=0;
    memset(dp,-1,sizeof dp);
    while(n)
    {
        num[p++]=n%10;
        n/=10;
    }
    LL ans=0;
    for(int i=0;i<=num[p-1];i++)
    {
        int tool=(i*reminder[p-1])%13;
        ans+=dfs(p-1,i,0,i!=num[p-1],(13-tool)%13);
    }
    return ans;
}
int main()
{
  int n;
  init();
  while(cin>>n)
  {
      cout<<solve(n)<<'\n';
  }
   return 0;
}