UESTC Windy Digital DP

Title Link: http://acm.uestc.edu.cn/#/problem/show/250

Title Description:

Number of WindyTime limit:3000/1000ms (java/others) Memory limit:65535/65535kb (java/others)SubmitStatus

Windy defines a windy number.

A positive integer that does not contain a leading 0 and the difference of at least 2 of the adjacent two digits is called the windy number.

Windy want to know, in A And B Between, including A And B How many windy are there in total?

Input

Contains two integers, A B 。

Meet 1≤A≤B≤2000000000 .

Outputsample Input and output

sample Input	sample Output
1	9

Sourcewindy Test Instructions: Title said, test instructions is very clear; thought: This is my study of digital DP, the first full self-thinking + code implementation of the problem; Dp[i][j][z]: The first I-digit, ending with J, Z==1 indicates that a leading is a 0,z==0, leading is not 0 Contains number of windy memory search ~ Of course, the table is OK ~

#include <iostream> #include <stdio.h> #include <string> #include <string.h>using namespace std ; int dp[25][25][3];int bit[25];int dfs (int pos,int pre,int flag,int z) {  if (pos==0) return 1;  if (!flag && dp[pos][pre][z]!=-1) return dp[pos][pre][z];  int end=flag?bit[pos]:9;  int ans=0;  for (int i=0;i<=end;i++)  {    if (z==1 | | pre-i>=2 | | i-pre>=2) ANS+=DFS (pos-1,i,flag&& (i==end), z&& (i==0));  }  if (!flag) Dp[pos][pre][z]=ans;  return ans;} int cal (int x) {int len=0; memset (bit,0,sizeof (bit)); Memset (Dp,-1,sizeof (DP)); while (x) {   bit[++len]=x%10;   x/=10; } return Dfs (len,0,1,1);} int main () {  int l,r;  while (scanf ("%d%d", &l,&r)!=eof)  {      int s1=cal (r);      int s2=cal (L-1);      printf ("%d\n", s1-s2);  }  return 0;}

UESTC Windy Digital DP