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 (≤10^4) 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. Notice that the first digit must not be zero.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

思路

• 贪心：对字符串 s1 与 s2，如果 s1 + s2 < s2 + s1，那么把 s1 放在 s2 的前面。

注意点

• 结果串的所有前导零需要去掉。由于每个字符串都是非负的整数（non-negative），是有可能为 0 的，所以如果去除前导零后串的长度变为0，则输出0。

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

const int maxn = 10000;

int cmp(string s1, string s2) {
    return s1 + s2 < s2 + s1;
}

int main() {
    int n;
    string str[maxn];
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> str[i];
    }
    sort(str, str + n, cmp);
    string ans;
    for (int i = 0; i < n; i++) {
        ans += str[i];
    }
    while (ans.size() != 0 && ans[0] == '0') {
        ans.erase(ans.begin());
    }
    if (ans.size() == 0) {
        cout << 0;
    } else {
        cout << ans;
    }
    return 0;
}