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。
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