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

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 (≤104) 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

题目大意：给出几个数，要求求出它们的片段拼接，并且这个数是所有数中最小的。

//一看到就不太明白怎么做，拼接不同总长度也不一定相同，有0开头的，如果放在中间的话就算是中间一位了。没什么思路，考试遇到这个的话会跪。

代码来自：https://www.liuchuo.net/archives/2303

#include <iostream>

#include <string>

#include <algorithm>

using namespace std;

bool cmp0(string a, string b) {//两个相同的数相加，它们的长度肯定是相同的。

    return a + b < b + a;

}

string str[];//使用字符串数组！

int main() {

    int n;

    scanf("%d", &n);

    for(int i = ; i < n; i++)

        cin >> str[i];//以String作为输入。

    sort(str, str + n, cmp0);

    string s;

    for(int i = ; i < n; i++)

        s += str[i];//将字符串拼接。

    while(s.length() !=  && s[] == '')

        s.erase(s.begin());//将开头的0除去。

    if(s.length() == ) cout << ;

    cout << s;

    return ;

}

//柳神的代码简直叹为观止。厉害了，学习了。

1.对字符串的处理，关键是这个cmp0函数的使用。

巴特西