PAT 甲级 1038 Recover the Smallest Number (30分)

题目

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N个数字，求出将这些数字拼接得到的数字中最小的那个数字

思路

当只有两个数字a、b的时候，有a拼接b和b拼接a两种情况，此时，将这两个数字用字符串表示的话，当a+b < b+a的时候，a+b表示的数字小于 b+a表示的数字；
当有N个数字当时候，对N个数字以此方法排序即可

代码

#include<bits/stdc++.h>
using namespace std;
bool cmp(const string a, const string b){
    return a+b < b+a;
}
int main(int argc, const char * argv[]) {
    int N;
    string s, res="";
    cin>>N;
    vector<string> v(N);
    for(int i=0; i<N; i++){
        cin>>v[i];
    }
    sort(v.begin(), v.end(), cmp);
    for(auto it : v)
        res += it;
    int st = 0;
    while(st < res.size() && res[st] == '0') st++;
    if(st == res.size()) cout<<'0';
    else{
        for(int i=st; i<res.size(); i++)
            cout<<res[i];
    }
    return 0;
}