题目传送门
1 /*
2 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。
3 另外，最多len-1次循环
4 */
5 #include <cstdio>
6 #include <algorithm>
7 #include <cstring>
8 #include <cmath>
9 using namespace std;
10
11 typedef long long ll;
12 const int MAXN = 1e3 + 10;
13 const int INF = 0x3f3f3f3f;
14
15 int get_len(ll x) {
16 int ret = 0;
17 while (x) {
18 x /= 10; ret++;
19 }
20 return ret;
21 }
22
23 int get_nine(ll x) {
24 int ret = 0;
25 while (x) {
26 ll y = x % 10; x /= 10;
27 if (y != 9) break;
28 ret++;
29 }
30 return ret;
31 }
32
33 int main(void) { //Codeforces Round #135 (Div. 2) B. Special Offer! Super Price 999 Bourles!
34 //freopen ("C.in", "r", stdin);
35
36 ll p, d;
37 while (scanf ("%I64d%I64d", &p, &d) == 2) {
38 int len = get_len (p); ll now = p;
39 int mx = get_nine (p); ll ans = p;
40 ll cut = 1;
41 while (true) {
42 ll y = now / cut % 10;
43 if (y != 9) {
44 now = now - cut * 10; now = now / cut + cut - 1;
45 }
46 cut *= 10;
47 if (now < p - d) break;
48 int cnt = get_nine (now);
49 if (cnt > mx) ans = now;
50 }
51
52 printf ("%I64d
", ans);
53 }
54
55 return 0;
56 }