思路：

数位dp

代码：

#include
     
      using namespace std;#define LL long long #define pb push_back#define mem(a, b) memset(a, b, sizeof(a))int a[15];int dp[15][15];int dfs(int pos, int pre, bool limit, bool zero) {    if (pos == -1) return 1;    if(!limit && !zero && ~dp[pos][pre]) return dp[pos][pre];    int top = limit ? a[pos] : 9;    int ans = 0;    for (int i = 0; i <= top; i++) {        if(zero){            ans += dfs(pos - 1, i, limit&&i==a[pos], zero&&i==0);        }        else {            if(abs(pre - i) <= 1) continue;            ans += dfs(pos - 1, i, limit&&i==a[pos], zero&&i==0);        }    }     if(!limit && !zero) dp[pos][pre] = ans;    return ans;}int solve(int n) {    int cnt = 0;    mem(a, 0);     while (n) {        a[cnt++] = n % 10;        n /= 10;    }    return dfs(9, 0, 1, 1);} int main() {    int a, b;    scanf("%d%d", &a, &b);    mem(dp, -1);    printf("%d\n",solve(b) - solve(a - 1));    return 0;}