1038. Recover the Smallest Number (30)【排序】——PAT (Advanced Level) Practise_1038 recover the smallest number 冒泡排序-优快云博客

本文探讨了如何通过排序算法解决最小组合数问题，详细介绍了输入格式、输出要求及解题策略，提供了AC代码实例。

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题目信息

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

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

解题思路

排序，注意排序的比较函数较奇特

AC代码

#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
string node[10005];
bool cmp(const string s1, const string s2){
    return s1 + s2 < s2 + s1;
}
int main()
{
    int n;
    char ts[10];
    scanf("%d", &n);
    for (int i = 0; i < n; ++i){
        scanf("%s", ts);
        node[i] = ts;
    }
    sort(node, node + n, cmp);
    string s;
    for (int i = 0; i < n; ++i){
        s += node[i];
    }
    n = 0;
    while (n < s.size() && s[n] == '0') ++n;
    if (n == s.size()){
        printf("0\n");
    }else{
        printf("%s\n", s.substr(n).c_str());
    }
    return 0;
}