[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 }

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