构造 Codeforces Round #135 (Div. 2) B. Special Offer! Super Price 999 Bourles!

 /*

    构造：从大到小构造，每一次都把最后不是9的变为9，p - p MOD 10^k - 1，直到小于最小值。

         另外，最多len-1次循环

 */

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 using namespace std;

 typedef long long ll;

 const int MAXN = 1e3 + ;

 const int INF = 0x3f3f3f3f;

 int get_len(ll x)   {

     int ret = ;

     while (x)   {

         x /= ;    ret++;

     }

     return ret;

 }

 int get_nine(ll x)  {

     int ret = ;

     while (x)   {

         ll y = x % ;  x /= ;

         if (y != ) break;

         ret++;

     }

     return ret;

 }

 int main(void)  {       //Codeforces Round #135 (Div. 2) B. Special Offer! Super Price 999 Bourles!

     //freopen ("C.in", "r", stdin);

     ll p, d;

     while (scanf ("%I64d%I64d", &p, &d) == )   {

         int len = get_len (p);  ll now = p;

         int mx = get_nine (p);  ll ans = p;

         ll cut = ;

         while (true)    {

             ll y = now / cut % ;

             if (y != ) {

                 now = now - cut * ;    now = now / cut + cut - ;

             }

             cut *= ;

             if (now < p - d)    break;

             int cnt = get_nine (now);

             if (cnt > mx)   ans = now;

         }

         printf ("%I64d\n", ans);

     }

     return ;

 }