题目传送门
1 /*
2 数学：不会写，学习一下这种解题方式:)
3 思路：设符合条件的数的最高位是h，最低位是l，中间不变的部分为mid，由题意可得到下面的公式(这里对X乘上1e6用a表示，b表示1e6)
4 (h*power+l+mid)*a = (l*power+h+mid)*b
5 可推得：mid = ((h*power+l) * a - (l*power+h) * b) / (a - b);
6 所以可以枚举h，l然后求mid，注意mid的最低位一定是0，因为留出最低位加l或者h
7 详细解释：http://blog.csdn.net/u010660276/article/details/46290703
8 */
9 #include <cstdio>
10 #include <algorithm>
11 #include <cstring>
12 #include <cmath>
13 #include <vector>
14 using namespace std;
15
16 typedef long long ll;
17 const int MAXN = 1e4 + 10;
18 const int INF = 0x3f3f3f3f;
19 vector<ll> V;
20
21 int main(void) //2015百度之星初赛2 HDOJ 5255 魔法因子
22 {
23 int t, cas = 0; scanf ("%d", &t);
24 while (t--)
25 {
26 V.clear ();
27 double x; scanf ("%lf", &x);
28 ll a = (ll) (x * 1e6 + 0.1); ll b = 1e6;
29
30 for (int k=2; k<=10; ++k)
31 {
32 ll p = 1; for (int i=1; i<k; ++i) p *= 10;
33 for (int r=1; r<=9; ++r)
34 {
35 for (int l=0; l<=9; ++l)
36 {
37 ll tmp = (l * p + r) * b - (r * p + l) * a;
38 if (tmp % (10 * (a - b)) == 0)
39 {
40 ll mid = tmp / 10 / (a - b); ll tmp_mid = mid;
41 if (tmp_mid < 0) continue;
42 int cnt = 0; while (tmp_mid) {++cnt; tmp_mid /= 10;}
43 if (cnt <= k - 2) V.push_back (r * p + mid * 10 + l);
44 }
45 }
46 }
47 }
48
49 printf ("Case #%d:
", ++cas);
50 printf ("%d
", V.size ());
51 for (int i=0; i<V.size (); ++i)
52 {
53 printf ("%I64d%c", V[i], (i == V.size ()-1) ? '
' : ' ');
54 }
55 }
56
57 return 0;
58 }
59
60
61 /*
62 3
63 3.1312
64 3.1215
65 0.3
66 */