PAT甲1038 Recover the smallest number

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 (≤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

 

题意：

给定n个数，要求任意组合使得组合得到的数最小。

思路：

之前做过一道类似的。这道题给的样例好啊。

表面上看好像是应该将输入的数字存成字符，按照字典序从小到大输出。但是观察样例会发现并不是这样的。

因为单个的字典序最小并不代表组合了之后的字典序最小。

应该要写一个cmp， 用的是组合之后字典序最小的排前面

还有要注意都是0的时候要记得最后输出一个0

#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f

const int maxn = 1e4 + 5;
string strnum[maxn];
int n;

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

int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        cin>>strnum[i];
    }
    //while(cin>>strnum[n++]);
    sort(strnum, strnum + n, cmp);
    bool flag = true;
    for(int i = 0; i < n; i++){
        int l = strnum[i].length();
        for(int j = 0; j < l; j++){
            if(flag && strnum[i][j] == '0')continue;
            if(flag && strnum[i][j] != '0'){
                flag = false;
            }
            printf("%c", strnum[i][j]);
        }
    }
    if(flag){
        printf("0");
    }
    printf("\n");
    return 0;
}