pat1038. Recover the Smallest Number (30)

1038. Recover the Smallest Number (30)

时间限制
400 ms
内存限制
65536 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

提交代码

学习网址:http://blog.csdn.net/sinat_29278271/article/details/48047877

里面的传递性是可以证明的

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 #include<cmath>
 8 #include<string>
 9 using namespace std;
10 vector<string> v;
11 bool cmp(string a,string b){
12     return a+b<b+a;
13 }
14 int main(){
15     //freopen("D:\INPUT.txt","r",stdin);
16     int n,i;
17     scanf("%d",&n);
18     string s;
19     for(i=0;i<n;i++){
20         cin>>s;
21         v.push_back(s);
22     }
23     sort(v.begin(),v.end(),cmp);
24     s="";
25     for(i=0;i<n;i++){
26         s=s+v[i];
27     }
28     for(i=0;i<s.length();i++){
29         if(s[i]!='0'){
30             break;
31         }
32     }
33 
34     if(i==s.length()){
35         printf("0
");
36     }
37     else{
38         for(;i<s.length();i++){
39             cout<<s[i];
40         }
41         cout<<endl;
42     }
43     return 0;
44 }
原文地址:https://www.cnblogs.com/Deribs4/p/4770206.html