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

分析：

输出几个数字字符串，输出其组合数的最小值，最简单的当然是穷举然后排序了，可是太暴力了，我们要反对暴力。

应该先对字符串进行一个排序。然后直接组合后输出就可以。这里的比較函数是一个难点。

注意到234<2343, 234 > 2341。所以假设两个字符串比較，较短字符串比較完后还要进行循环比較直到较长字符的末尾（即要取%接着比較）

要注意输入都为0的情况。也注意一下string的erase的使用方法

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string>
#include<vector>
using namespace std;
vector<string>V;
bool cmp1(string A, string B){//核心的比較函数
 if (A.size() > B.size()){
  for (int i = 0; i < A.size(); i++){
   if (A[i] != B[i % B.size()])//当较短字符串比較完了之后。还要循环继续比較到较长字符串的末尾，由于234是小于2343的。可是是大于2341的。
    return A[i] < B[i % B.size()];
  }
 }
 else{
  for (int i = 0; i < B.size(); i++){
   if (A[i % A.size()] != B[i])
    return A[i % A.size()] < B[i];
  }
 }//make sure that A is no longer than B
}
int main(){
 //freopen("F://Temp/input.txt", "r", stdin);
 string tmp;
 int n;
 cin >> n;
 for (int i = 0; i < n; i++){
  cin >> tmp;
  V.push_back(tmp);
 }
 sort(V.begin(), V.end(), cmp1);
 string str;
 for (int i = 0; i < V.size(); i++){
  str += V[i];
 }
 int idx = 0;
 while (str[idx] == '0')idx++;
 str.erase(0, idx);//要把首部的0给去除掉
 if (str.size() == 0)cout << "0" << endl;//假设一个数字也没有，输出0
 else cout << str << endl;
 return 0;
}

截图：

——Apie陈小旭