PAT A1038 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

30分的题目，之前想了好久都没思路，本来想的是通过整型数组一个个匹配取最小的，但是写起来显然错误。偷瞄了一眼提示，想起来可以用STL中的字符串来做！由于这是我第一次做string有关知识的题，边学边做，踩到的坑记录如下（还是太菜了）。

1. #include< string > 而非#include< string.h >。
2. 读入整个字符串，用cin （读出用cout或c_str()转为字符数组后printf)
3. a+b可以实现直接将b拼接到a后。
4. 得到字符串长度直接用a.length()。另：strlen是char*型字符串，需要先c_str()后使用。

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
bool cmp(string a,string b){
	return a+b<b+a;
}
int main(){
	int N;
	cin>>N;
	string s[10010];
	for(int i=0;i<N;i++)
	cin>>s[i];
	sort(s,s+N,cmp);
	string ans;
	for(int i=0;i<N;i++){
		ans=ans+s[i];
	}
	int L=ans.length();
	int flag=0;
	int n=0;     //第一个非0的位置 
	if(ans[0]=='0') flag=1;
	if(flag==1){            //开头是0就要把0去掉 
		for(int i=1;i<L;i++){
			if(ans[i]!='0'){
				n=i;
				break;
			}
		}
		if(n==0){        //全部都是0只输出0就好
			printf("0");
			return 0;
		}	
	}
	for(int i=n;i<L;i++){     //其实可以erase删除0元素后直接输出
		printf("%c",ans[i]);
	}
	return 0;
}