http://pat.zju.edu.cn/contests/pat-practise/1038

本文链接：https://blog.youkuaiyun.com/cstur4/article/details/8066684

本文介绍了一个算法挑战，即从给定的数字段集合中找出能够组成最小数字的组合方式。通过对数字段进行特定的排序和组合，确保生成的数字是最小的。文章提供了一个C++实现示例，通过字符串拼接的方式比较大小，并最终输出结果。

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1038. Recover the Smallest Number (30)

     时间限制
   

     400 ms
    
 
   

     内存限制
   

     32000 kB
    
 
   

     代码长度限制
   

     16000 B
    
 
   

     判题程序
   
 
    Standard
   

    作者
  

    CHEN, Yue
   
 
  
 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 (<=10000) 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. Do not output leading zeros.
Sample Input:5 32 321 3214 0229 87
Sample Output:22932132143287

   
 
  
[cpp]view plaincopy 
      
 #include <iostream>  
 #include <string>  
 #include <algorithm>  
 #include <memory.h>  
 #include <cstdio>  
 #include <cstdlib>  
 #include <vector>  
 using namespace std;  
 #define MAX 0Xfffffff  
 string a[10001];  
   
 bool cmp(string s1, string s2){  
       
     return s1+s2<s2+s1;  
 }  
 int main(){  
   
     //freopen("in.txt", "r", stdin);  
   
     int n;  
     cin>>n;  
     for(int i=0;i<n;++i){  
         cin>>a[i];  
     }  
     sort(a, a+n, cmp);  
     bool first = true;  
     bool print = false;  
     for(int i=0;i<n;++i){  
         int len = a[i].length();  
         for(int j=0;j<len;++j){  
             if(first){  
                 if(a[i][j]!='0'){  
                     printf("%c", a[i][j]);  
                     print = true;  
                     first = false;  
                 }         
             }else{  
                 printf("%c", a[i][j]);  
                 print = true;  
             }  
         }  
     }  
     if(!print)  
         printf("0");  
     printf("\n");  
     //fclose(stdin);  
 }     