HDU 3652 数位DP 解题报告

B-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

【解题报告】

这是机房模拟ACM的A题，然而只有我一个人做了。。。

关于不能出现13的部分就非常显然了，重点就在整除上面，因为如果直接判断的话肯定是没法转移的，所以大概思路就是记录这个数除13的余数，（DP多加一维，dfs的时候加几句就可以了）。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n;
int dp[10][15][2][10],a[10];

int dfs(int pos,int mod,int t,int pre,int limit)
{
if(pos==-1)return (mod==0&&t);
if(!limit&&dp[pos][mod][t][pre]!=-1)return dp[pos][mod][t][pre];
int up=limit?a[pos]:9;
int ans=0;
for(int i=0;i<=up;++i)
{
ans+=dfs(pos-1,(mod*10+i)%13,t||(pre==1&&i==3),i,limit&&(i==up));
}
if(!limit) dp[pos][mod][t][pre]=ans;
return ans;
}
int solve(int x)
{
int len=0;
while(x)
a[len++]=x%10,x/=10;
return dfs(len-1,0,0,0,1);
}

int main()
{
memset(dp,-1,sizeof(dp));
while(scanf("%d",&n)!=EOF)
{
printf("%d\n",solve(n));
}
return 0;
}