hdu3652 B-number（数位dp）
题目传送门
B-number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8970 Accepted Submission(s): 5331
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
Recommend
题意：求[1,n]中数位有13并且能被13整除的数
题解：数位dp，状态就是要记录上一位，还要记录这个数，可开3维或者4维
代码：
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
int n;
//int dp[10][2][13];
int dp[10][10][2][13];//i:处理的数位，j:结尾的数 k:是否已经包含13 z:该数对13取模以后的值，
int bit[10];
int dfs(int pos,int pre,bool flag,int sta,bool limit)
{
if(pos==-1)return sta%13==0&&flag;
if(!limit&&dp[pos][pre][flag][sta%13]!=-1)return dp[pos][pre][flag][sta%13];
int up=limit?bit[pos]:9;
int ans=0;
for(int i=0;i<=up;i++)
{
/*if(pre==2||pre==1&&i==3)
ans+=dfs(pos-1,2,sta*10+i,limit&&i==up);
else if(i==1)
ans+=dfs(pos-1,1,sta*10+i,limit&&i==up);
else
ans+=dfs(pos-1,0,sta*10+i,limit&&i==up);*/
ans+=dfs(pos-1,i,flag||(pre==1&&i==3),sta*10+i,limit&&i==up);
}
if(!limit)dp[pos][pre][flag][sta%13]=ans;
return ans;
}
int calc(int x)
{
int len=0;
while(x)
{
bit[len++]=x%10;
x/=10;
}
return dfs(len-1,-1,0,0,1);
}
int main()
{
memset(dp,-1,sizeof(dp));
while(~scanf("%d",&n))
{
printf("%d\n",calc(n));
}
return 0;
}
