1038 Recover the Smallest Number （30 分）

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。

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-26-11.49.45
 * Description : A1038
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;
 ;
 string a[maxn];
 bool cmp(string a,string b){
     return a+b<b+a;  //如果a+b<b+a，把a+b排在前面
 }
 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     int n;
     cin>>n;
     ;i<n;i++){
         cin>>a[i];
     }
     sort(a,a+n,cmp);
     string ans="";
     ;i<n;i++){
         ans+=a[i];
     }
     ]=='){
         ans.erase(ans.begin());
     }
     if(ans.size()) cout<<ans;
     ";
     ;
 }