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。

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 int n, m, cnt, a[];
 ll dp[][][];

 ll dfs(int pos, int pre, int state0, int limit) {
     if (!pos)    return ;
     if (!limit && dp[pos][pre][state0] >= )    return dp[pos][pre][state0];

     int high = limit ? a[pos] : ;
     ll ret = ;
     for (int low = ; low <= high; low++) {
         if (abs(low - pre) <  && !state0)    continue;
         ret += dfs(pos - , low, (low == ) & state0, (low == a[pos]) & limit);
     }

     if (!limit)    dp[pos][pre][state0] = ret;
     return ret;
 }

 ll solve(int x) {
     for (cnt = ; x; x /= )
         a[++cnt] = x % ;
     return dfs(cnt, -, , );
 }

 int main() {
     memset(dp, -, sizeof dp);
     cin >> n >> m;
     cout << solve(m) - solve(n - ) << endl;
     return ;
 }