[csu1392]YY一下

题意：给定x，求有多少个10^8以内的数满足这个数乘以x以后，最高位到了最低位。设最高位的数字和剩余长度，列等式推理即可。

 1 #pragma comment(linker, "/STACK:10240000,10240000")
 2
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cstdlib>
 7 #include <cstring>
 8 #include <map>
 9 #include <queue>
10 #include <deque>
11 #include <cmath>
12 #include <vector>
13 #include <ctime>
14 #include <cctype>
15 #include <set>
16
17 using namespace std;
18
19 #define mem0(a) memset(a, 0, sizeof(a))
20 #define lson l, m, rt << 1
21 #define rson m + 1, r, rt << 1 | 1
22 #define define_m int m = (l + r) >> 1
23 #define Rep(a, b) for(int a = 0; a < b; a++)
24 #define lowbit(x) ((x) & (-(x)))
25 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
26 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
27 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
28
29 typedef double db;
30 typedef long long LL;
31 typedef pair<int, int> pii;
32 typedef multiset<int> msi;
33 typedef multiset<int>::iterator msii;
34 typedef set<int> si;
35 typedef set<int>::iterator sii;
36 typedef vector<int> vi;
37
38 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
39 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1};
40 const int maxn = 1e5 + 7;
41 const int maxm = 1e5 + 7;
42 const int maxv = 1e7 + 7;
43 const int MD = 1e9 +7;
44 const int INF = 1e9 + 7;
45 const double PI = acos(-1.0);
46 const double eps = 1e-10;
47
48 int digit(LL x) {
49     int cnt = 0;
50     while (x) {
51         cnt++;
52         x /= 10;
53     }
54     return cnt;
55 }
56
57 int main() {
58     //freopen("in.txt", "r", stdin);
59     double tx;
60     while (cin >> tx) {
61         LL x = (int)(tx * 10000 + 0.5), get = 0;
62         if (x >= 100000) {
63             puts("No solution");
64             continue;
65         }
66         LL p = 1;
67         for (int i = 0; i <= 7; i++) {
68             for (int k = 1; k <= 9; k++) {
69                 LL tmp = k * (x * p - 1e4);
70                 if (tmp % (LL)(1e5 - x)) continue;
71                 tmp /= 1e5 - x;
72                 if (digit(tmp) == i) {
73                     printf("%d", k);
74                     if (tmp > 0) printf("%d", tmp);
75                     puts("");
76                     get = 1;
77                 }
78             }
79             p *= 10;
80         }
81         if (!get) puts("No solution");
82     }
83     return 0;
84 }