1038. Recover the Smallest Number (30)

Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given {32, 321, 3214, 0229, 87}, we can recover many numbers such like 32-321-3214-0229-87 or 0229-32-87-321-3214 with respect to different orders of combinations of these segments, and the smallest number is 0229-321-3214-32-87.

Input Specification:

Each input file contains one test case. Each case gives a positive integer N (<=10000) followed by N number segments. Each segment contains a non-negative integer of no more than 8 digits. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the smallest number in one line. Do not output leading zeros.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

来源： <http://www.patest.cn/contests/pat-a-practise/1038>

 

1. #include <iostream>
2. #include <stdio.h>
3. #include <vector>
4. #include <algorithm>
5. #include <string>
6. #pragma warning(disable:4996)
8. using namespace std;
10. struct Num {
11. string s;
12. char c[8];
13. };
15. vector<Num> num;
17. bool cmp(Num a, Num b) {
18. for (int i = 0; i < 8; i++) {
19. if (a.c[i] < b.c[i]) {
20. return true;
21. }
22. else if(a.c[i] > b.c[i]){
23. return false;
24. }
25. }
26. return true;
27. }
29. bool CheckInput(string s) {
30. for (int i = 0; i < s.length(); i++) {
31. if (s[i] != '0')
32. return true;
33. }
34. return false;
35. }
37. int main(void) {
38. int n;
39. cin >> n;
40. /*char chartemp;
41. scanf("%c", &chartemp);*/
42. Num numtemp;
43. for (int i = 0; i < n; i++) {
44. cin >> numtemp.s;
45. if (!CheckInput(numtemp.s))
46. continue;
47. for (int i = 0; i < 8; i++) {
48. numtemp.c[i] = numtemp.s[i % (numtemp.s.length())];
49. }
50. num.push_back(numtemp);
51. }
54. if (num.size() == 0) {
55. cout << "0";
56. return 0;
57. }
59. sort(num.begin(), num.end(), cmp);
61. for (int i = 0; i < num.size(); i++) {
62. if (i == 0) {
63. bool flag = false;
64. for (int j = 0; j < num[i].s.length(); j++) {
65. if (num[i].s[j] != '0')
66. flag = true;
67. if (flag == true)
68. cout << num[i].s[j];
69. }
70. }
71. else
72. cout << num[i].s;
73. }
74. return 0;
75. }

来自为知笔记(Wiz)

转载于:https://www.cnblogs.com/zzandliz/p/5023132.html