PAT 1038. Recover the Smallest Number (30) 字符串大小比较

1038. Recover the Smallest Number (30)

时间限制

400 ms

内存限制

65536 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

想了半天没想明白怎么判断两个字符串的排序关系，比如 578和5781这种。看了人家的比较，拍案叫绝。

附上我的代码

#include<string>#include<stdio.h>#include<iostream>#include<string.h>#include<queue>#include<algorithm>using namespace std;  bool cmp1(string x,string y)  {      string xy="";      string yx="";      xy+=x;      xy+=y;      yx+=y;      yx+=x;      return xy<yx;  }  int main(){    int n;    cin>>n;    string aa[10005];    for(int i=0;i<n;i++)      cin>>aa[i];      sort(aa,aa+n,cmp1);      string last="";      for(int i=0;i<n;i++)        last+=aa[i];        int p=0;        while(last[p]=='0') p++;        if(p==last.size()) cout<<'0';        else        for(;p<last.size();p++)        cout<<last[p];       return 0;}