因为最多17位数字 且dfs的递归方程最多只有2种
因此最多操作次数为2^17 = 131,072 足够
code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int a, b;
ll n;
ll f[20];
int cnt = 1;
ll mx;

void dfs(int no, ll ans){
    if(no == cnt){
        mx = max(mx, ans);
        return;
    }

    int up = 9 - f[no], down = f[no] + 1;
    if(up <= a){
        a -= up;
        dfs(no + 1, ans * 10 + 9);
        a += up;
    }
    else{
        ll t = a;
        a = 0;
        dfs(no + 1, ans * 10 + t + f[no]);
        a = t;
    }
    if(down <= b){
        b -= down;
        dfs(no + 1, ans * 10 + 9);
        b += down;
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> a >> b;
    ll sum = n;

    while(sum){
        f[cnt++] = sum % 10;
        sum /= 10;
    }
    reverse(f + 1, f + cnt);
    dfs(1, 0);
    cout << mx;
    return 0;
}