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[11];
ll dp[11][10][2];

ll dfs(int pos, int pre, int state0, int limit) {
    if (!pos)    return 1;
    if (!limit && dp[pos][pre][state0] >= 0)    return dp[pos][pre][state0];

    int high = limit ? a[pos] : 9;
    ll ret = 0;
    for (int low = 0; low <= high; low++) {
        if (abs(low - pre) < 2 && !state0)    continue;
        ret += dfs(pos - 1, low, (low == 0) & state0, (low == a[pos]) & limit);
    }

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

ll solve(int x) {
    for (cnt = 0; x; x /= 10)
        a[++cnt] = x % 10;
    return dfs(cnt, -1, 1, 1);
}

int main() {
    memset(dp, -1, sizeof dp);
    cin >> n >> m;
    cout << solve(m) - solve(n - 1) << endl;
    return 0;
}