windy数，数位dp

中文题面:http://www.lydsy.com/JudgeOnline/problem.php?id=1026

用着这样的数位dp姿势搞下，就好了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define INF 1000000000


int dp[20][20][2];

int pos[20];
int judge(int x, int y)
{
	if (abs(x - y) >= 2) return 1;
	return 0;
}

int nong(int x, int last, int first, int flag)
{
	int ans = 0;
	if (x == 0) return 1;
	if (!flag&&~dp[x][last][first]) return dp[x][last][first];
	int bound = flag ? pos[x] : 9;
	for (int i = 0; i <= bound; i++){
		if (first == 0){
			ans += nong(x - 1, i, first || i, flag&&i == bound);
		}
		else{
			if (judge(i, last)) ans += nong(x - 1, i, first || i, flag&&i == bound);
		}
	}
	if (!flag) dp[x][last][first] = ans;
	return ans;
}

int gao(int x)
{
	int len = 0;
	while (x){
		pos[++len] = x % 10;
		x /= 10;
	}
	return nong(len, 0, 0, 1);
}


int main()
{
	int A, B;
	memset(dp, -1, sizeof(dp));
	while (cin >> A >> B){
		cout << gao(B) - gao(A - 1) << endl;
	}
	return 0;
}