1038 Recover the Smallest Number (30 分)

Posted 2022-03-26 zllwxm123

1038 Recover the Smallest Number (30 分)

 

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 (≤) 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. Notice that the first digit must not be zero.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

这题挺有意思的，一开始我还没用这种排序，用那种字符排序，
有点蠢，仔细想一下，这个就是要求最小，所以拼接比较是最直观的。

 1 #include <bits/stdc++.h>
 2 #define N 10054
 3 using namespace std;
 4 int n;
 5 struct Node
 6     string s;
 7     friend bool operator <(const Node a, const Node b)
 8         return a.s+b.s < b.s+a.s;
 9     
10 ;
11 Node an[N];
12 
13 int main()
14     cin >> n;
15     for(int i = 0 ; i < n; i++)
16         cin >> an[i].s;
17     
18     sort(an,an+n);
19     bool flag = false;
20     for(int i = 0; i < n; i++)
21         for(int j = 0 ; j < an[i].s.length(); j++)
22             if(an[i].s[j]!=‘0‘ || flag)
23                 cout <<an[i].s[j];
24                 flag = true;
25             
26     
27     if(!flag)
28         cout <<"0";
29     cout << endl;
30     return 0;
31