bzoj1026: [SCOI2009]windy数 数位dp

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int N = 30 + 10;
 6 
 7 int dp[N][10];//定义dp[i][j]为第i位前驱数字为j时的方案数，注意前驱要合法
 8 int dig[N];
 9 int tot = 0;
10 int dfs(int pos, int pre, bool f, bool limit)
11 {//f用来标记是否有合法前驱
12     if(pos < 1) return 1;
13     if(!limit && f && dp[pos][pre] != -1) return dp[pos][pre];
14     int en = limit ? dig[pos] : 9;
15     int ans = 0;
16     for(int i = 0; i <= en; i++)
17         if(! f) ans += dfs(pos-1, i, f || i != 0, limit && i == en);
18         else if(abs(i - pre) >= 2) ans += dfs(pos-1, i, f || i != 0, limit && i == en);
19     if(! limit && f) dp[pos][pre] = ans;
20     return ans;
21 }
22 int work(int n)
23 {
24     int tn = n;
25     tot = 0;
26     while(tn) dig[++tot] = tn % 10, tn /= 10;
27     memset(dp, -1, sizeof dp);
28     return dfs(tot, 0, 0, 1);
29 }
30 int main()
31 {
32     int n, m;
33     while(~ scanf("%d%d", &n, &m))
34         printf("%d\n", work(m) - work(n-1));
35     return 0;
36 }