"Problem solution": windy number

Question: windy number

Time limit: 1 Sec   Memory Limit: 512 MB

Face questions

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Title Description

It defines a number of Windy Windy: without leading zeros and adjacent to at least the difference between two numbers

The number of positive integers is called Windy.

Windy want to know, in A and B between, including A and B, a total of how many Windy number?

Input Format

Two line numbers, respectively, A, B .

Output Format

Output An integer that represents the answer.

Sample input

1 10

Sample Output

9

answer

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My Digital dp entry title, ah, in fact, quite easy.

Set f [i] [j] denotes the i fill bits, the maximum number is the number of bits of windy j.

So do not consider what God hath pilot 0 of issues to consider temporary position gap of at least 2 issues directly and vigorously dp calculated all the values.

Then do it again vigorously dp, j from 1-9 cycles accumulate answer it.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define int long long
#define rint register int
#define ll long long
using namespace std;
int f[20][15]={0},a,b;
ll pw[15];
void prework()
{
    pw[0]=1;
    for(rint i=1;i<=12;++i)pw[i]=pw[i-1]*10;
    for (RINT I = 0 ; I <= . 9 ; I ++) F [ . 1 ] [I] = . 1 ;
     for (I = RINT 2 ; I <= . 11 ; I ++) // enumerator bits 
        for (RINT = J 0 ; J <= . 9 ; J ++) // enumeration highest 
            for (RINT K = 0 ; K <= . 9 ; K ++) // a most significant bit of the state enumeration, views this state high 
                IF (ABS (JK)> = 2 ) F [I] [J] + = F [I- . 1 ] [K]; 
} 
int COUNT ( int X) 
{ 
    int W = 0 , Y, ANS =0,pre;
    while(pw[w]<=x)++w;//求位数
    for(rint i=1;i<w;++i)//枚举位数
        for(rint j=1;j<=9;++j)//枚举最高位
            ans+=f[i][j];
    y=x/pw[w-1];
    for(rint i=1;i<y;++i)ans+=f[w][i];
    pre=y;
    x%=pw[w-1];
    for(rint i=w-1;i>=1;--i)
    {
        y=x/pw[i-1];
        for(rint j=0;j<y;++j)
            if(abs(j-pre)>=2)
                ans+=f[i][j];
        if(abs(pre-y)<2)break;
        pre=y;
        x%=pw[i-1];
    }
    return ans;
}
signed main()
{
    scanf("%lld %lld",&a,&b);
    prework();
    cout<<count(b+1)-count(a)<<endl;
    return 0;
}

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