hdu3652B-number（数位dp）

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6444 Accepted Submission(s):
3739


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
Process till EOF. In each line, there is one positive
integer n(1 <= n <= 1000000000).

[align=left]Output[/align]
Print each answer in a single line.

[align=left]Sample Input[/align]

13
100
200
1000

[align=left]Sample Output[/align]

1
1
2
2

[align=left]Author[/align]
wqb0039

[align=left]Source[/align]
2010
Asia Regional Chengdu Site —— Online Contest

题意：找出1~n范围内含有13并且能被13整除的数字的个数

思路:http://blog.csdn.net/libin56842/article/details/10026063

#include<cstdio>
#include<iostream>
#include<cstring>

using namespace std;
int dit[15],f[15][15][3];

int dfs(int pos,int mod,int have,int lim)
{
int num,ans,mod_x,have_x;
if(pos<=0) return mod==0 && have==2;
if(!lim && f[pos][mod][have]!=-1) return f[pos][mod][have];
num=lim?dit[pos]:9;ans=0;
for(int i=0;i<=num;i++)
{
mod_x=(mod*10+i)%13; have_x=have;
if(have==0 && i==1) have_x=1;
if(have==1 && i!=1) have_x=0;
if(have==1 && i==3) have_x=2;
ans+=dfs(pos-1,mod_x,have_x,lim&&i==num);
}
if(!lim) f[pos][mod][have]=ans;
return ans;
}

int main()
{
int n,len;
while(~scanf("%d",&n))
{
memset(dit,0,sizeof dit);
memset(f,-1,sizeof f);len=0;
while(n)
{
dit[++len]=n%10;
n/=10;
}dit[len+1]=0;
printf("%d\n",dfs(len,0,0,1));
}
return 0;
}