PAT 甲级 刷题日记|A 1038 Recover the

单词积累

recover 恢复；弥补；重新获得

题目

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
结尾无空行

Sample Output:

22932132143287
结尾无空行

思路

这道题最开始处理得太麻烦了，得分还少。经过学习，发现最关键的是sort函数的使用，对str1和str2的排序，与其详细的指定规则，不如直接str1+str2和str2+str1大小比较返回，非常巧妙啊！

此外，还要注意，组合形成的数首位不能为0。

代码

#include <bits/stdc++.h>
using namespace std;

const int maxn = 10003;
string num[maxn];

bool cmp (string str1, string str2) {
    return str1 + str2 < str2 + str1;
}

int main() {
    int N;
    cin>>N;
    for (int i = 0; i < N; i++) {
        cin>>num[i];
    } 
    sort(num, num + N, cmp);
    
    string ans;
    for (int i = 0; i < N; i++) {
        ans += num[i];
    } 
    while(ans.size() != 0 && ans[0] == '0'){
        ans.erase(ans.begin());
    }
    if(ans.size() == 0) printf("0\n");
    else cout << ans<<endl;

}