1038 Recover the Smallest Number （30 分）

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

題目大意：給出幾個數，要求求出它們的片段拼接，並且這個數是所有數中最小的。

//一看到就不太明白怎麽做，拼接不同總長度也不一定相同，有0開頭的，如果放在中間的話就算是中間一位了。沒什麽思路，考試遇到這個的話會跪。

代碼來自：https://www.liuchuo.net/archives/2303

#include <iostream>
#include <string>
#include <algorithm>
using
 
 namespace std;
bool cmp0(string a, string b) {//兩個相同的數相加，它們的長度肯定是相同的。
    return a + b < b + a;
}
string str[10010];//使用字符串數組！
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        cin >> str[i];//以String作為輸入。
    sort(str, str + n, cmp0);
    string s;
    for(int i = 0; i < n; i++)
        s += str[i];//將字符串拼接。
    while(s.length() != 0 && s[0] == ‘0‘)
        s.erase(s.begin());//將開頭的0除去。
    if(s.length() == 0) cout << 0;
    cout << s;
    return 0;
}

//柳神的代碼簡直嘆為觀止。厲害了，學習了。

1.對字符串的處理，關鍵是這個cmp0函數的使用。

PAT 1038 Recover the Smallest Number[dp][難]