BZOJ1026（数位dp）

数位dp第二道～就当成搜索，套板子写了写。我写的dp[pos][pre][state0]记录的是当前pos位没有限制时、前面的数是pre时、前面是否都是0时的方案数。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 typedef long long ll;
 8 int n, m, cnt, a[11];
 9 ll dp[11][10][2];
10 
11 ll dfs(int pos, int pre, int state0, int limit) {
12     if (!pos)    return 1;
13     if (!limit && dp[pos][pre][state0] >= 0)    return dp[pos][pre][state0];
14 
15     int high = limit ? a[pos] : 9;
16     ll ret = 0;
17     for (int low = 0; low <= high; low++) {
18         if (abs(low - pre) < 2 && !state0)    continue;
19         ret += dfs(pos - 1, low, (low == 0) & state0, (low == a[pos]) & limit);
20     }
21 
22     if (!limit)    dp[pos][pre][state0] = ret;
23     return ret;
24 }
25 
26 ll solve(int x) {
27     for (cnt = 0; x; x /= 10)
28         a[++cnt] = x % 10;
29     return dfs(cnt, -1, 1, 1);
30 }
31 
32 int main() {
33     memset(dp, -1, sizeof dp);
34     cin >> n >> m;
35     cout << solve(m) - solve(n - 1) << endl;
36     return 0;
37 }