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 (≤10?4??) 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


题意：输入n个非负数数，求这n个数字组成的最小数

分析：贪心，对各个字符串排序，如果直接从小到大排序，会出现错误，比如32和321，排序后合并为32321，显然32132比它更小。

注意到这点题目就好做了，对sort自定义比较函数，a+b<b+a，那么让a+b在前面即可。

有个测试点是去掉前导0的时候可能字符串长度会变成0，此时直接输出0。

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-26-11.49.45
 6 * Description : A1038
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=10010;
20 string a[maxn];
21 bool cmp(string a,string b){
22     return a+b<b+a;  //如果a+b<b+a，把a+b排在前面
23 }
24 int main(){
25 #ifdef ONLINE_JUDGE
26 #else
27     freopen("1.txt", "r", stdin);
28 #endif
29     int n;
30     cin>>n;
31     for(int i=0;i<n;i++){
32         cin>>a[i];
33     }
34     sort(a,a+n,cmp);
35     string ans="";
36     for(int i=0;i<n;i++){
37         ans+=a[i];
38     }
39     while(ans[0]==‘0‘){
40         ans.erase(ans.begin());
41     }
42     if(ans.size()) cout<<ans;
43     else cout<<"0";
44     return 0;
45 }