本文共 1529 字，大约阅读时间需要 5 分钟。

1038 Recover the Smallest Number （30 point(s)）

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

Experiential Summing-up

这一题....其实弄懂了原理，就不难了，排序的规则一定得弄懂，比如  01230 和 0123  前者就应该排在前面，因为3后面接的是0，确定比后面出现以1为开头的字符串要小，按照顺序排好之后，剩下的就是从小到大依次组成答案，然后去掉开头多余的0就行啦，这里再举一个例子，045，12，这两个组成的最小数是4512，可不是12045~  以后一定注意，如果无法AC，多自己想想例子，如果自己举的例子都想不通，那么说明你的思路有问题，得好好检查一下= =

Accepted Code

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;bool cmp(string a,string b){ int i; for(i=0;i
         
          
           b[0];}int main(){ int n; scanf("%d",&n); string a; vector
           
             element[10]; for(int i=0;i
            
             >a; element[a[0]-'0'].push_back(a); } for(int i=0;i<10;++i) sort(element[i].begin(),element[i].end(),cmp); string answer; for(int i=0;i<10;++i) for(int j=0;j
            
           
          
         
        
       
      
     
    
   

转载地址：https://meteorrain.blog.csdn.net/article/details/86655646 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！