HDU 3652 （数位dp）

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7127    Accepted Submission(s): 4166

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

131002001000

 

Sample Output

1122
题意： 你的任务是找出含有 13 并且能被 13整除的数的个数。
 dp[i][mod][type][pre]：i:处理的数位，mod:该数对13取模以后的值，type:是否已经包含13,pre结尾的数 
代码：
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dp[22][15][2][10];int num[22];int dfs(int pos,int mod,int type,int pre,int lim){//type 为 1  表示含有 13 为 0 表示不含有 13  if(pos<=0){if(type&&mod==0) return 1;//到最低位 如果mod==0 并且 有13存在  那么久返回 1。 else return 0;}if(!lim&&dp[pos][mod][type][pre]!=-1) return dp[pos][mod][type][pre];int end=lim?num[pos]:9;int fin=0;for(int i=0;i<=end;i++){fin+=dfs(pos-1,(mod*10+i)%13,type||(pre==1&&i==3),i,lim&&i==end);//这里的（mod*10)%13  其实就是一个除法的模拟。直到除到最后一位，//如果mod为0  那么表示这个数可以被13整除 }if(!lim) dp[pos][mod][type][pre]=fin;return fin;}void solve(int x){int len=0;while(x){num[++len]=x%10;x/=10;}int fin=dfs(len,0,0,0,1);printf("%d\n",fin);return ;}int main(){int n;memset(dp,-1,sizeof(dp));while(scanf("%d",&n)!=EOF){solve(n);}return 0;}