轩辕互动Exoweb笔试题（一）

1        定义一个数组中，若某个值出现的次数在一半以上，则为执行者。如下数组： A[8]={1，2，2，4，2，4，2，2} 由于2出现上午次数5次大于5/8>0.5， 所以2为数组A[8]的执行者。

解法一：

/****************************************************************

* Function: Find the number who appears over half of the total times in an array

*

* @Author: Zheng Haishu    Date: 2009-12-31

****************************************************************/

#include<iostream>

using namespace std;

 

void Administer(int find[], int length)

{

      int i,j;

      int found = 0;  // 0 indicates not found the administer

                            // 1 indicates found the administer

      int num = 0;

      for(i=0; i<length&&!found; i++)

      {

           num = find[i];

           int count = 0;

           for(j=0; j<length; j++)

           {

                 if(num == find[j])

                      count++;

           }

          

           if(count>length/2)

           {

                 cout<<num<<" is an administer"<<endl;

                 found = 1; // if found administer, break the loop

           }

      }

}

 

int main(void)

{

      int array[] = {1,2,2,4,2,4,2,2};

    Administer(array, 8);

      return 0;

}

运行结果：

2 is an administer

 

 

 

解法二：

/****************************************************************

* Function: Find the number who appears over half of the total times in an array

*

* @Author: Zheng Haishu    Date: 2009-12-31

****************************************************************/

#include<iostream>

using namespace std;

 

int Administer(int find[], int length)

{

      int number;

      int nTimes=0;

      int i;

      for(i = 0; i < length; i++)

      {

           if(nTimes == 0)

           {

                 number = find[i];

                 nTimes = 1;

           }

           else

           {

                 if(number == find[i])

                      nTimes++;

                 else

                      nTimes--;

           }

      }

      return number;

}

 

int main(void)

{

      int array[] = {1,2,2,4,2,4,2,2};

    int administer = Administer(array, 8);

      cout<<administer<<" is an administer."<<endl;

      return 0;

}

 

运行结果：

2 is an administer

 

 

2        平衡点：比如int[] numbers = {1,3,5,7,8,25,4,20}; 25前面的总和为24，25后面的总和也是24，25这个点就是平衡点；假如一个数组中的元素，其前面的部分等于后面的部分，那么这个点的位序就是平衡点。要求：返回任何一个平衡点

 

解：

#include<iostream>

using namespace std;

 

void Equilibrium(int find[], int length)

{

      int i,j,k;

      int lSum; // sum of left side

      int rSum; // sum of right side

      for(i=0; i<length; i++)

      {

           lSum=0;

           rSum=0;

           for(j=0;j<i;j++)

           {

                 lSum += find[j];

           }

           for(k=i+1; k<length; k++)

           {

                 rSum += find[k];

           }

           if(lSum == rSum)

           {

                 cout<<find[i]<<" is an equilibrium point"<<endl;    

           }

      }

}

 

int main(void)

{

      int array[]={1,3,5,7,8,25,4,20};

      Equilibrium(array, 8);

      return 0;

}

运行结果：

 

25 is an equilibrium point

 

 

3        实现一个去除整型数组中绝对值相同的数字.最后返回一个长度

#include<iostream>

using namespace std;

 

void NewLength(int find[], int length)

{

      int i,j;

      for(i=0; i<length; i++)

      {

           if(find[i] == 0)

                 continue;

           for(j=i+1; j<length; j++)

           {

                 if(find[j] == find[i] || find[j] == (-find[i]))

                 {

                      find[i] = 0;

                      find[j] = 0;

                      break;

                 }

           }

      }

      int newLength = 0;

      for(i=0; i<length; i++)

      {

           if(find[i] != 0)

           {

                 cout<<find[i]<<" ";

                 newLength++;

           }

      }

      cout<<endl;

      cout<<"Length of new array is "<<newLength<<endl;

 

}

 

int main(void)

{

      int array[] = {1,2,2,3,-4,4,5,6,7,-6};

    NewLength(array, 10);

      return 0;

}

 

运行结果：

1 3 5 7

Length of new array is 4

 

 

4   找排序后的值，在原数组中的下标

#include<iostream>

using namespace std;

#define length 8

 

void SortAndFind(int find[])

{

      int i,j,temp;

      int originArray[length];

      for(i=0; i<length; i++)

      {

           originArray[i] = find[i];

      }

 

    for(j=1; j<=length-1; j++)

      {

           for(i=0;i<=length-1-j;i++)

           {

                 if(find[i] > find[i+1])

                 {

                      temp = find[i];

                      find[i] = find[i+1];

                      find[i+1] = temp;

                 }

           }

      }

 

      int number;

      cout<<"Input a number: ";

      cin>>number;

      for(i=0; i<length; i++)

      {

           if(number == originArray[i])

           {

                 cout<<"The index of "<<number<<" is "<<i<<endl;

           }

      }

 

}

 

int main(void)

{

      int array[] = {4,3,2,8,1,5,7,6};

    SortAndFind(array);

      return 0;

}

运行结果：

Input a number: 4

The index of 4 is 0