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

题目给出一组数据片段，求其可以组成的最小的数字并输出。

解题思路

1. 读入所有数据片段，并排序，排序依据为使a+b<b-a（贪心法）；
2. 将排好序的数据片段连接起来，并去除首位0，直到其首位非零或为空；
3. 如结果非空输出结果，否则输出“0”（边界条件）。
4. 返回零值。

代码

#include<iostream>
#include<algorithm>
#include<string>
#define maxn 10010
using namespace std;
bool cmp(string a,string b){
    return a+b<b+a;
}

int main(){
    int i,N;
    string a[maxn],sum;
    cin>>N;
    sum.clear();
    for(i=0;i<N;i++){
        cin>>a[i];
    }
    sort(a,a+N,cmp);
    for(i=0;i<N;i++){
        sum+=a[i];
    }
    while(sum.size()!=0&&sum[0]=='0'){
        sum.erase(sum.begin());
    }
    if(sum.size()==0){
        cout<<"0"<<endl;
    }else{
        cout<<sum<<endl;
    }
    return 0;
}

运行结果