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

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/75356219

本文介绍了一个算法问题，即如何从给定的多个数字段中构造出最小可能的数字。通过特定的排序策略来确保组合后的数字最小，并给出了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题意：给你n个数，将它们按一定顺序连起来，使得连起来后最小

解题思路：关键在于a+b<b+a，根据这个排序，然后还是注意下前导零

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

string s[10009];
int n;

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

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