uva 10905 - Children's Game

4^thIIUCInter-University Programming Contest, 2005

A

Children’s Game

Input: standard input
Output: standard output

Problemsetter: Md. Kamruzzaman

There are lots of number games for children. These games are pretty easy to play but not so easy to make. We will discuss about an interesting game here. Each player will be givenN positive integer. (S)He can make a big integer by appending those integers after one another. Such as if there are 4 integers as 123, 124, 56, 90 then the following integers can be made – 1231245690, 1241235690, 5612312490, 9012312456, 9056124123 etc. In fact 24 such integers can be made. But one thing is sure that 9056124123 is the largest possible integer which can be made.

You may think that it’s very easy to find out the answer but will it be easy for a child who has just got the idea of number?

Input

Each input starts with a positive integer N (≤ 50). In next lines there areN positive integers. Input is terminated by N = 0, which should not be processed.

Output

For each input set, you have to print the largest possible integer which can be made by appending all theN integers.

Sample Input

Output for Sample Input

4
123 124 56 90
5
123 124 56 90 9
5
9 9 9 9 9
0

9056124123
99056124123
99999

 

直接按单个数字assic大小排序再连接会导致错误是因为，对于前缀相同长度不等的数字无法判断如何连接，对于前缀不同或者长度相同的可以直接连接

如

98       9 98 979   ASSIC  98>979>9
979
9 

123    123 12 1    ASSIC 123>12>1 
12 
1 

因此对所有两两连接的assic排序,可以保证前缀大的再前面而且连接之后大的也在前面

#include<string.h>#include<stdio.h>int main(){ int i,j,n; char s[51][100],ch1[200],ch2[200],temp[100]; while (scanf("%d",&n),n) {  for (i=1;i<=n;i++)      scanf("%s",&s[i]);  for (i=1;i<n;i++)  for (j=i+1;j<=n;j++)  {   strcpy(ch1,s[i]);   strcpy(ch2,s[j]);   strcat(ch1,s[j]);   strcat(ch2,s[i]);   if (strcmp(ch1,ch2)<0) {strcpy(temp,s[i]); strcpy(s[i],s[j]); strcpy(s[j],temp);}  }  for (i=1;i<=n;i++)  printf("%s",s[i]);  printf("\n"); } return 0;}