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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本文介绍了一个算法问题，即从给定的一系列数字片段中构造出可能的最小数值。通过两两比较数字片段组合的方式确定最优顺序，并最终形成最小数字。

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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

思路：两两相加进行比较，ab和ba比较如果，哪个小的话，第一个字符串要在前面，最后结合起来就是最小的结果。

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>

using namespace std;

bool comp(const string &s1, const string &s2) 
{
	return (s1 + s2) < (s2 + s1);
}
int main()
{
	int n;
	cin >> n;
	vector <string> vs;
	string s;
	while (n--) {
		cin >> s;
		vs.push_back(s);
	}
	string ans;
	sort(vs.begin(), vs.end(), comp);
	for (int i=0; i!=vs.size(); ++i) 
		ans += vs[i];
	int i;
    for (i=0; i!=ans.size(); ++i) //去掉前面的0
		if (ans[i] != '0')
			break;
	if (i == ans.size())
		cout << 0 << endl;
	else 
		cout << ans.substr(i) << endl;
	return 0;
}