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c语言面试编程题
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;
}
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