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c语言面试编程题

发布时间:2019-05-15

  1、读文件 file1.txt 的内容(例如):

  12

  34

  56

  输出到 file2.txt:

  56

  34

  12

  #include

  #include

  int main(void)

  {

  int max = 10;

  int *a = (int *)malloc(max * sizeof(int));

  int *b;

  file *fp1;

  file *fp2;

  fp1 = fopen("a.txt","r");

  if(fp1 == null)

  {printf("error1");

  exit(-1);

  }

  fp2 = fopen("b.txt","w");

  if(fp2 == null)

  {printf("error2");

  exit(-1);

  }

  int i = 0;

  int j = 0;

  while(fscanf(fp1,"%d",&a[i]) != eof)

  {

  i++;

  j++;

  if(i >= max)

  {

  max = 2 * max;

  b = (int*)realloc(a,max * sizeof(int));

  if(b == null)

  {

  printf("error3");

  exit(-1);

  }

  a = b;

  }

  }

  for(;--j >= 0;)

  fprintf(fp2,"%d\n",a[j]);

  fclose(fp1);

  fclose(fp2);

  return 0;

  }

  2、写一段程序,找出数组中第 k 大小的数,输出数所在的位置。例如{2,4,3,4,7}中,第一大的数是 7,位置在 4。第二大、第三大的数都是 4,位置在 1、3 随便输出哪一个均可。

  函数接口为:int find_orderk(const int* narry,const int n,const int k)

  要求算法复杂度不能是 o(n^2)

  可以先用快速排序进行排序,其中用另外一个进行地址查找代码如下,在 vc++6.0 运行通过。

  //快速排序

  #include

  usingnamespacestd;

  intpartition (int*l,intlow,int high)

  {

  inttemp = l[low];

  intpt = l[low];

  while (low < high)

  {

  while (low < high && l[high] >= pt)

  --high;

  l[low] = l[high];

  while (low < high && l[low] <= pt)

  ++low;

  l[low] = temp;

  }

  l[low] = temp;

  returnlow;

  }

  voidqsort (int*l,intlow,int high)

  {

  if (low < high)

  {

  intpl = partition (l,low,high);

  qsort (l,low,pl - 1);

  qsort (l,pl + 1,high);

  }

  }

  intmain ()

  {

  intnarry[100],addr[100];

  intsum = 1,t;

  cout << "input number:" << endl;

  cin >> t;

  while (t != -1)

  {

  narry[sum] = t;

  addr[sum - 1] = t;

  sum++;

  cin >> t;

  }

  sum -= 1;

  qsort (narry,1,sum);

  for (int i = 1; i <= sum;i++)

  cout << narry[i] << '\t';

  cout << endl;

  intk;

  cout << "please input place you want:" << endl;

  cin >> k;

  intaa = 1;

  intkk = 0;

  for (;;)

  {

  if (aa == k)

  break;

  if (narry[kk] != narry[kk + 1])

  {

  aa += 1;

  kk++;

  }

  }

  cout << "the no." << k << "number is:" << narry[sum - kk] << endl;

  cout << "and it's place is:" ;

  for (i = 0;i < sum;i++)

  {

  if (addr[i] == narry[sum - kk])

  cout << i << '\t';

  }

  return0;

  }

来源:www.diyifanwen.com/qiuzhijianli/msshiti/469335.htm

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