PAT_A 1038 Recover the Smallest Number

PAT_A 1038 Recover the Smallest Number

分析

题意为将N个数字相连接成为一个新数字，求新数字最小的连接情况。

可通过排序将输入序列转换为一个新序列。显然，对于两数字a、b及连接运算 +，若 a+b < b+a 则a应排在b之前，推广至整个序列仍成立。因此排序后按序输出即可。

需要注意最后合成的新数字是不含前导0的。因此需要对于含有多个 0 元素（例如0, 00等）的输入；以及全0的输入做相应处理。

PAT_A 1038 Recover the Smallest Number

题目的描述

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

AC的代码

#include<bits/stdc++.h>

using namespace std;

bool cmp(string a,string b){
    return stoll(a+b)<stoll(b+a);
}

int main(){
    int N;
    scanf("%d",&N);
    vector<string> num(N);
    for(int i=0;i<N;i++){
        cin>>num[i];
    }
    sort(num.begin(),num.end(),cmp);
    bool f=false;
    int j=0;
    while(!f && j<N){
        for(int i=0;i<num[j].length();i++){
            if(num[j][i]=='0'&&!f)continue;
            cout<<num[j][i];
            f=true;
        }
        j++;
    }

    for(int i=j;i<N;i++){
        cout<<num[i];
        f=true;
    }
    if(!f)cout<<'0';
    cout<<endl;
    return 0;
}