PAT-1038

1038. Recover the Smallest Number 

题目地址: http://pat.zju.edu.cn/contests/pat-practise/1038

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

排序题目,排序的原则:
<1>在长度相同的情况,ASCII码表的顺序排列
<2>长度不相等时,假设len(a) > len(b), 逐次比较a+len(b),和b两个字符串在min(a,b),长度上的大小 

逐次比较的过程可以采用递归或者是循环,递归看上去简洁,这里就采用了递归的写法


#if 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
#define MIN(a,b) ((a) < (b) ? (a):(b))

int cmpstr(char *a,char *b,int n)
{
     int alen = strlen(a);
     int blen = strlen(b);
     if(strncmp(a,b,n) == 0)
            return cmpstr(a+n,b,n);
     else
            return strncmp(a,b,n);
}

int cmp(const void *a,const void *b)
{
     int alen = strlen(( char*)a);
     int blen = strlen(( char*)b);
     int min = MIN(alen,blen);
     if(min == blen)
            return cmpstr(( char*)a,( char *)b,min);
     else
            return cmpstr(( char*)b,( char *)a,min) < 0 ? 1 : -1;
}

#define MAXN  (100000+1)

char a[MAXN][40];
int main(int argc,char *argv[])
{
     int i,j;
     int n;
     int flag;
     freopen( "D:\\Chengsq\\in.txt" , "r" ,stdin);
     freopen( "D:\\Chengsq\\out.txt", "w" ,stdout);

     while(scanf( "%d",&n) != EOF)
     {
            for(i = 0;i < n;i++)
                scanf( "%s",a[i]);


           qsort(a,n, sizeof(a[0]),cmp);
           flag = 0;
            for(i = 0;i < n;i++)
           {
                 for(j = 0; j < strlen(a[i]);j++)
                {
                      if(flag)           //是否已经输出非零的字符  
                           printf( "%c",a[i][j]);
                      else
                            if(a[i][j] != '0')
                           {
                                printf( "%c",a[i][j]);  //输出第一个非零的字符
                                flag = 1;
                           }


                }
           }

            if(flag ==0)
                printf( "0");
           printf( "\n");
     }
     return 0;
}

#endif


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