1038 Recover the Smallest Number （30 分)

2019-03-17

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:

1	5 32 321 3214 0229 87

Sample Output:

1	22932132143287

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

拼接一串数字，使之最小。

分析

一开始想的是类似桶排序，比较数字各位的大小，选小的放在前面。但是有点难实现。后来看到了使用cmp函数的解法，非常巧妙。有a，b两个字符串，比较a+b和b+a的大小，再利用sort函数就可以了。最后注意去除字符串开头的0。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;

bool cmp(string s1,string s2){
    return s1+s2<s2+s1;
}
int main() {
    int n;
    cin>>n;
    vector<string> v;
    string s;
    for(int i=0;i<n;i++) {
        cin >> s;
        v.push_back(s);
    }
    sort(v.begin(),v.end(),cmp);
    string res="";
    for(string s:v)
        res+=s;
    int i;
    for(i=0;i<res.length();i++)
        if(res[i]!='0')
            break;
        if(i==res.length())
            cout<<"0"<<endl;
        else{
            for(;i<res.length();i++)
                cout<<res[i];
        }
}