PAT (Advanced Level) Practise 1038 Recover the Smallest Number (30)

1038. Recover the Smallest Number (30)

時間限制

400 ms

記憶體限制

65536 kB

代碼長度限制

16000 B

判題程式

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

排個序，去掉前導零就好了。排序的關鍵是a+b<b+a

#include<cstdio>
#include<string>
#include<stack>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;

const int maxn = 1e5 + 10;
typedef long long LL;
int n;
string s[maxn];

bool cmp(const string& a, const string& b)
{
  return a + b < b + a;
}

int main()
{
  scanf("%d", &n);
  for (int i = 0; i < n; i++) cin >> s[i];
  sort(s, s + n, cmp);
  int flag = 1;
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; s[i][j]; j++)
    {
      if (flag&&s[i][j] == '0') continue;
      flag = 0; printf("%c", s[i][j]);
    }
  }
  if (flag) printf("0\n");
  return 0;
}