Children's Game UVA - 10905 (简单贪心)

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 given

N

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 nd 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 are

N

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

the

N

integers.

Sample Input

4

123 124 56 90

5

123 124 56 90 9

5

9 9 9 9 9

0

Sample Output

9056124123

99056124123

99999

点拨：注意贪心策略 可利用字符串进行编写。

 

Code：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int n;
string arr[105];
int cmp(string a,string b){
	int len1=a.length();
	int len2=b.length();
 	while(len1 && len2){
 		
	 	if(len1>len2){
		 	for(int i=0;i<len2;i++){
			 	if(a[i]!=b[i]) return a[i]>b[i];
			 }
			a=a.substr(len2);
		 }
		 else{
		 	for(int i=0;i<len1;i++){
			 	if(a[i]!=b[i]) return a[i]>b[i];
			 	
			 }
			 b=b.substr(len1);
		 }
		 len1=a.length();
		 len2=b.length();
	 }
	 return 1;
}
int main(){
	while(cin>>n && n){
		for(int i=0;i<n;i++) cin>>arr[i];
		sort(arr,arr+n,cmp);
		for(int i=0;i<n;i++) cout<<arr[i]; cout<<endl;
	
	} 
	return 0; 
}