PAT 1038 体验Python之美

1038. Recover the Smallest Number (30)

时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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 (<=10000) 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. Do not output leading zeros.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

Python代码

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print int(''.join(sorted(raw_input().split()[1:], lambda s1, s2: cmp(s1+s2, s2+s1))))

是的只有一行!可惜最后一个test还是超时了。于是用C++再写:

C++代码

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#include <set>
#include <sstream>
#include <string>
#include <iostream>
using namespace std;

class mstring : public string {
public:
inline bool operator<(const string &s) const {
string tmpa(*this);
tmpa.append(s);
string tmpb(s);
tmpb.append(*this);
return tmpa.compare(tmpb) < 0;
}
};

int main() {
//#pragma warning(disable:4996)
//freopen("..\\advanced-pat-python\\test.txt", "r", stdin);
int N;
cin >> N;
set<mstring> nums;
mstring ms;
while (N--) {
cin >> ms;
nums.insert(ms);
}
mstring num;
set<mstring>::iterator it = nums.begin();
while (it!=nums.end()) {
num += *(it++);
}
while (!num.empty() && *num.begin()=='0') {
num.erase(num.begin());
}
if (num.empty()) {
num += '0';
}
cout << num << endl;
}

总体来说还是较为简单的,但是没有Python那么爽的一行解决掉了。Python真是神奇的语言啊。