1038 Recover the Smallest Number

1038 Recover the Smallest Number （30 分)

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 (≤10​4​​) 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. Notice that the first digit must not be zero.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

解析：题意就不讲了。这题真的搞了好久，后面实在想不到点上，看了网上的解析“一点即通”，网上的大佬还说这是一道面试的题目，只是改动了。题目中要求组成的数要最小，有若a+b<b+a;说明a在b前面是比较小的，我们给个符号“<<”定义，a<<b是a恒小于b，"<<"有递推性，a<<b,b<<c,则a<<c;

这样直接就一个排序解决了问题。

坑点：样例

1.0000   1111

2.0000   0000

注意处理上面的两个样例

#include<bits/stdc++.h>
using namespace std;

#define ee exp(1)
#define p acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}


string a[10005];
int cmp(string x,string y)
{
	return x+y<y+x;
}
int main()
{
	int n;cin>>n;
	for(int i=0; i<n; i++)
	{
		cin>>a[i];
	}
	sort(a,a+n,cmp);
	
	
	bool flag=true;
	for(int i=0; i<n; i++)
	{
		if(i==0)
		{
			for(int j=0; j<a[0].size(); j++)
			if(a[0][j]=='0'&&flag)continue;
			else flag=false,cout<<a[0][j];
		}
		else
		{
			if(flag)
			{
				for(int j=0; j<a[i].size(); j++)
				if(a[i][j]=='0'&&flag)continue;
				else flag=false,cout<<a[i][j];
			}
			else cout<<a[i];
		}
	}
	if(flag)
	puts("0");
	return 0;
}