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

Posted 2021-01-08 bluebluesea

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

 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[10010];//使用字符串数组！
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        cin >> str[i];//以String作为输入。
    sort(str, str + n, cmp0);
    string s;
    for(int i = 0; i < n; i++)
        s += str[i];//将字符串拼接。
    while(s.length() != 0 && s[0] == ‘0‘)
        s.erase(s.begin());//将开头的0除去。
    if(s.length() == 0) cout << 0;
    cout << s;
    return 0;
}

 

//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。