1038. Recover the Smallest Number (30)(贪心)

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本文深入探讨了编程技术的核心概念及其在实际应用中的实现，通过具体实例展示了如何解决复杂编程难题，旨在帮助开发者提升技术能力，增强解决实际问题的能力。

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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

题解:

这题采取贪心的策略

比如说一个字符串abcdes1s2xyz 与 abcdes2s1xyz这两个字符串的大小仅仅与s1s2 和 s2s1的大小有关系,所以, 我们可以在sort里边加个比较函数,最后输出就好了.

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

using namespace std;

int convert_to_int(const string& str) {
    int x = 0;
    for (int i = 0; i < str.length(); ++i)
        x = x * 10 + str[i] - '0';
    return x;
}

bool cmp(const string& s1, const string& s2) {
    return s1 + s2 < s2 + s1;
}

int main() {
    int n;
    scanf("%d", &n);
    vector<string> v;
    string str;

    for (int i = 0; i < n; ++i) {
        cin >> str;
        if (convert_to_int(str) > 0)
            v.push_back(str);
    }

    sort(v.begin(), v.end(), cmp);
    if (v.empty())
        puts("0");
    else {
        printf("%d", convert_to_int(v[0]));
        for (int i = 1; i < v.size(); ++i)
            cout << v[i];
        cout << endl;
    }

    return 0;
}