1038. Recover the Smallest Number (30)

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最新推荐文章于 2025-01-13 15:16:10 发布

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分类专栏： PAT习题

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本文介绍了一个算法挑战，即从给定的数字段集合中重构出可能的最小数值。通过重新定义字符串比较规则并使用排序算法，可以有效地解决这个问题。文章提供了两段不同的实现代码，并附带了输入输出示例。

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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

自己写的有一个测试点过不了，这道题的关键在于重新定义字符串的比较规则。

#include<iostream>
#include<fstream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>

#define inf 1000000
using namespace std;
bool Comp(string a, string b)
{
	
	int len1 = a.size(), len2 = b.size();
	if (len1 == len2)return a<b;
	else return a+b<b+a;
}
int main()
{
	
	int N;
	cin >> N;
	vector<string>list(N);
	int i;
	for (i = 0; i < N; i++)
		cin >> list[i];
	sort(list.begin(), list.end(), Comp);
	string sum;
	for (i = 0; i < list.size(); i++)
		sum += list[i];
	while (sum[0] == '0')
		sum.erase(sum.begin());
	cout << sum;
		
		return 0;
	
}

# include <cstdio>  
# include <iostream>  
# include <string>  
#include<algorithm>
using namespace std;  
  
const int _size = 10000;  
string num[_size];  
bool cmp(const string& a,const string& b){return a + b< b + a;}  
int main()  
{  
  int n,i;  
  cin >> n;  
  for (i=0;i<n;i++)  
     cin >> num[i];  
  sort(num,num+n,cmp);  
  string out;  
  for (i=0;i<n;i++)  
      out += num[i];  
  for (i=0;i<out.size()&&out[i]=='0';i++);  
  if (i==out.size())  
      printf("0");  
  else   
      printf("%s",out.c_str()+i);  
  printf("\n");  
  return 0;  
}