PAT甲1038 Recover the smallest number

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：

给定n个数，要求任意组合使得组合得到的数最小。

思路：

之前做过一道类似的。这道题给的样例好啊。

表面上看好像是应该将输入的数字存成字符，按照字典序从小到大输出。但是观察样例会发现并不是这样的。

因为单个的字典序最小并不代表组合了之后的字典序最小。

应该要写一个cmp， 用的是组合之后字典序最小的排前面

还有要注意都是0的时候要记得最后输出一个0

 #include <iostream>
 #include <set>
 #include <cmath>
 #include <stdio.h>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <map>
 using namespace std;
 typedef long long LL;
 #define inf 0x7f7f7f7f

 const int maxn = 1e4 + ;
 string strnum[maxn];
 int n;

 bool cmp(string a, string b)
 {
     return a + b < b + a;
 }

 int main()
 {
     scanf("%d", &n);
     for(int i = ; i < n; i++){
         cin>>strnum[i];
     }
     //while(cin>>strnum[n++]);
     sort(strnum, strnum + n, cmp);
     bool flag = true;
     for(int i = ; i < n; i++){
         int l = strnum[i].length();
         for(int j = ; j < l; j++){
             if(flag && strnum[i][j] == '')continue;
             if(flag && strnum[i][j] != ''){
                 flag = false;
             }
             printf("%c", strnum[i][j]);
         }
     }
     if(flag){
         printf("");
     }
     printf("\n");
     return ;
 }