【PAT1038】Recover the Smallest Number 字符串组合大小

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

比较 字符串a和字符串b的大小，只需要比较 ab < ba !!!

代码：

#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
using namespace std;

ifstream fin("in.txt");
#define cin fin

int cmp(const string& A,const string& B)
{
	string ab = A+B;
	string ba = B+A;
	return ab<ba;
}

int main()
{
	int n;
	cin>>n;
	int i;
	string* num = new string[n];
   for(i=0;i<n;i++)
   {
		cin>>num[i];
	}
   sort(num,num+n,cmp);
	string ss;
	for(i=0;i<n;i++)
		ss = ss+num[i];
	const char* ch = ss.c_str();
	i=0;
	while(ch[i]=='0')
	{
		i++;
	}
	ch = ch + i;
	if(ch[0]=='\0')cout<<0<<endl;
	else cout<<ch<<endl;
    system("PAUSE");
    return 0;
}