#include<bits/stdc++.h>
int n, m, a, k;
bool check(int f) {
	int t = 0, b = a;
	for (int i = 0; i < n; i ++) {
		int x = f % 10, y = b % 10;
		if (abs(x - y) == 1 || abs(x - y) == 9) t ++;
		else if (abs(x - y) >= 2) return 0;
		f /= 10, b /= 10;
	}
	return t <= k;
}
int main() {
	scanf("%d%d%d", &n, &a, &k);
	for (int i = 0; i < pow(10, n); i ++)
		if (check(i)) printf("%d\n", i);
	return 0;
}

最简单

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> run(int n, int a, int k) {
	vector<int> res;
	int g = a % 10;
	int b1 = g - 1, b2 = g + 1;
	if (b1 < 0) b1 = 9;
	else if (b2 > 9) b2 = 0;
	if (k <= 0) {
		res.push_back(a);
		return res;
	} else if (n == 1) {
		res.push_back(b1);
		res.push_back(b2);
		res.push_back(g);
		return res;
	}
	vector<int> l1 = run(n - 1, a / 10, k - 1);
	for (int i : l1)
		res.push_back(i * 10 + b1),
		res.push_back(i * 10 + b2);
	vector<int> l2 = run(n - 1, a / 10, k);
	for (int i : l2)
		res.push_back(i * 10 + g);
	return res;
}
int main() {
    int n, a, k;
    scanf("%d%d%d", &n, &a, &k);
    vector<int> resl = run(n, a, k);
    sort(resl.begin(), resl.end());
    for (int i : resl) printf("%d\n", i);
    return 0;
}

最优解