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

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 /*
    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 ;
 }