1038. Recover the Smallest Number (30)

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 (<=10000) 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. Do not output leading zeros.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

算法解析：采用贪心算法，把各个数字两两组合，按照组合起来的数字由小到大排序，然后再看顺序输出。这里要注意的是首位不能是0，除非要输出的数就是0。最关键的就是组合如何排序，只要用cmp()函数就能实现，具体的见代码部分。

#include <stdio.h>#include <stdlib.h>#include <string.h>int cmp(const void *a, const void *b) //构造比较组合大小的比较器{    char ab[20] = {'\0'}, ba[20] = {'\0'};    strcpy(ab, (char*)a);    strcpy(ba, (char*)b);    strcat(ab, (char*)b);    strcat(ba, (char*)a);    return strcmp(ab, ba);}int main(){    int n, i, j;    char collection[10000][9] = {{'\0'}};    scanf("%d", &n);    for(i = 0; i < n; i++)        scanf(" %s", collection[i]);    qsort(collection, n, sizeof(collection[0]), cmp); //实现排序    int flag = 1;    for(i = 0; i < n; i++)    {        int len = strlen(collection[i]);        for(j = 0; j < len; j++)        {            if(collection[i][j] == '0' && flag == 1) //如果第首位输出的数字是0，那么就不输出                continue;                        //此处代码可以化简，但为了逻辑清晰，还是这么写            else //当输出第一个不为0的数，即保证了首位不为0，就把flag清零。            {                printf("%c", collection[i][j]);                flag = 0;            }        }    }    if(flag == 1) //如果得到的数为0，就输出0        printf("0");    return 0;}