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 }
原文地址：https://www.cnblogs.com/Mered1th/p/10436431.html