Codeforces 489C - Given Length and Sum of Digits...（贪心）

题意

构造m位和为sum的最小和最大的数。

思路

一开始用了DFS，果断跪了。

贪心一下，如果要求最大的数，能填9的就填9，不够的补0，对于最小的数，先在第一位填1，然后从后面开始填9，不够补0，如果到了第一位有多的就全部加上去。

代码

1. #include
2. #include
3. #include
4. #include
5. #include
6. #include
7. #include
8. #include
9. #include
10. #include
11. #include
12. #include
13. #include
14. #include
15. #define LL long long
16. #define ULL unsigned long long
17. #define SZ(x) (int)x.size()
18. #define Lowbit(x) ((x) & (-x))
19. #define MP(a, b) make_pair(a, b)
20. #define MS(arr, num) memset(arr, num, sizeof(arr))
21. #define PB push_back
22. #define F first
23. #define S second
24. #define ROP freopen("input.txt", "r", stdin);
25. #define MID(a, b) (a + ((b - a) >> 1))
26. #define LC rt << 1, l, mid
27. #define RC rt << 1|1, mid + 1, r
28. #define LRT rt << 1
29. #define RRT rt << 1|1
30. #define BitCount(x) __builtin_popcount(x)
31. #define BitCountll(x) __builtin_popcountll(x)
32. #define LeftPos(x) 32 - __builtin_clz(x) - 1
33. #define LeftPosll(x) 64 - __builtin_clzll(x) - 1
34. const double PI = acos(-1.0);
35. const int INF = 0x3f3f3f3f;
36. using namespace std;
37. const double eps = 1e-8;
38. const int MAXN = 1000 + 10;
39. const int MOD = 1000007;
40. const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
41. typedef pair<int, int> pii;
42. typedef vector::iterator viti;
43. typedef vector::iterator vitii;
45. int minAns[MAXN], maxAns[MAXN];
47. int main()
48. {
49. int m, sum, i, j;
50. scanf("%d%d", &m, &sum);
51. int leave = sum;
52. if (m == 1 && sum == 0)
53. {
54. puts("0 0");
55. return 0;
56. }
57. if (9 * m < sum || sum == 0)
58. {
59. puts("-1 -1");
60. return 0;
61. }
62. minAns[0] = 1;
63. leave--;
64. for (int i = m - 1; i > 0; i--)
65. {
66. if (leave <= 9)
67. {
68. minAns[i] = leave;
69. leave = 0;
70. }
71. else
72. {
73. minAns[i] = 9;
74. leave -= 9;
75. }
76. }
77. minAns[0] += leave;
78. leave = sum;
79. for (int i = 0; i < m; i++)
80. {
81. if (leave <= 9)
82. {
83. maxAns[i] = leave;
84. leave = 0;
85. }
86. else
87. {
88. maxAns[i] = 9;
89. leave -= 9;
90. }
91. }
92. for (i = 0; i < m; i++) printf("%d", minAns[i]);
93. printf(" ");
94. for (i = 0; i < m; i++) printf("%d", maxAns[i]);
95. return 0;
96. }