去掉最高分和最低分求平均分数的问题

在百度知道看到个问题，原地址不记得了。题目大致是5位选手参加某歌唱比赛，然后6个评委依次为每位选手打分，最终要求出每位选手在去掉一个最高分和最低分后，自己的平均成绩是多少。然后有人给出了一个比较巧妙的答案，遍历二维数组中的每个元素的时候，“找出”最大值和最小值，然后就去掉这个最大值和最小值求出平均值了。

/**
 * 歌唱比赛评委评分，去掉最高分和最低分后求出平均分
 * @Filename:  singerscore2.c
 * @Source  :  百度知道
 */

#include <stdio.h>

int main( int argc, char * argv[] )
{
	//所有选手的评分
	float SoreData[5][6] = 
	{
		{9.31, 9.20, 9.00, 9.40, 9.35, 9.20},
		{9.71, 9.52, 9.50, 9.66, 9.49, 9.57},
		{8.89, 8.80, 9.10, 9.25, 8.90, 9.00},
		{9.38, 9.50, 9.40, 9.20, 9.90, 8.90},
		{9.30, 8.84, 9.40, 9.45, 9.10, 8.89}
	};
	
    float max, min, sum, average;
    int i, j;
    for( i=0; i<sizeof(SoreData)/sizeof(SoreData[0]); i++ )
    {
    	//初始化每个选手的总分数, 最低分, 最高分
        sum = 0; min = 10; max = 0;
        
        for( j=0; j<6; j++)
        {
            if( SoreData[i][j] > max )
                max = SoreData[i][j];
            if( SoreData[i][j] < min )
            	min = SoreData[i][j];
            sum += SoreData[i][j];
        }
        average = (sum-min-max)/4;
        
		//打印出结果
		printf( "No.[%d] total scores are:[%f], average scores are:[%.2f]\n\n", i+1, sum-min-max, (sum-min-max)/4 );
    }
    
    return 0;
}

测试结果：

No.[1] total scores are:[37.060003], average scores are:[9.27]

No.[2] total scores are:[38.249997], average scores are:[9.56]

No.[3] total scores are:[35.890002], average scores are:[8.97]

No.[4] total scores are:[37.480000], average scores are:[9.37]

No.[5] total scores are:[36.689996], average scores are:[9.17]

去掉最高分和最低分后求出平均分

自己想了想，还可以用冒泡排序法：先将二维数组中的每一行（即“一维数组”）用冒泡排序法排序，然后每一行中的起始元素和最后一个元素都不用管（分别是最小值、最大值），直接求平均值就行了。实现代码如下：

/**
 * 歌唱比赛评委评分，去掉最高分和最低分后求出平均分
 * @Filename:  singerscore.c
 * @author  :  vfhky 2014.11.10 https://typecodes.com
 */

#include <stdio.h>

//一维字符数组的冒泡排序
void BubbleSort( float array[], int n )
{
	int i=0, j=0; 
	float temp=0;
	int flag = 0;
	for( ; i<n-1; i++ )
	{
		flag = 0;
		for( j=n-1; j>i; j-- )
		{
			if( array[j] < array[j-1] )
			{
				temp = array[j];
				array[j] = array[j-1];
				array[j-1] = temp;
				flag = 1;
			}
		}
		
		if( flag == 0 )
			break;
	}
}


int main( int argc, char * argv[] )
{
	int i, j;
	
	//所有选手的评分
	float doubleSoreData[5][6] = 
	{
		{9.31, 9.20, 9.00, 9.40, 9.35, 9.20},
		{9.71, 9.52, 9.50, 9.66, 9.49, 9.57},
		{8.89, 8.80, 9.10, 9.25, 8.90, 9.00},
		{9.38, 9.50, 9.40, 9.20, 9.90, 8.90},
		{9.30, 8.84, 9.40, 9.45, 9.10, 8.89}
	};
	
	//二维数组冒泡排序
	for( i=0; i<5; i++ )
	{
		BubbleSort( doubleSoreData[i], 6 );
	}
	
	//初始化一个数组指针
	float (*p)[6];
	p = doubleSoreData;
	
	//声明每个选手的平均分
	float averagescore;
	
	//循环读取数据
	for( i=0; i<5; i++ )
	{
		//初始化每一个选手的平均分
		averagescore = 0;
		
		//除去最高/低位的数据
		for( j=1; j<5; j++ )
		{
			/**
			 * 利用数组读出数据
			 * printf( "doubleSoreData[%d][%d]=[%f]\n", i, j, doubleSoreData[i][j] );
			 * averagescore += doubleSoreData[i][j];
			 */
			
			//利用数组指针读出数据
			printf( "*((p+%d)[%d])=[%.2f]  ", i, j, (*(p+i))[j] );
			averagescore += (*(p+i))[j];
		}
		
		//打印出结果
		printf( "\nNo.[%d] total scores are:[%f], average scores are:[%.2f]\n\n", i+1, averagescore, averagescore/4 );
	}
	
	return 0;
}

测试结果：

*((p+0)[1])=[9.20]  *((p+0)[2])=[9.20]  *((p+0)[3])=[9.31]  *((p+0)[4])=[9.35]  
No.[1] total scores are:[37.059998], average scores are:[9.26]

*((p+1)[1])=[9.50]  *((p+1)[2])=[9.52]  *((p+1)[3])=[9.57]  *((p+1)[4])=[9.66]  
No.[2] total scores are:[38.250000], average scores are:[9.56]

*((p+2)[1])=[8.89]  *((p+2)[2])=[8.90]  *((p+2)[3])=[9.00]  *((p+2)[4])=[9.10]  
No.[3] total scores are:[35.889999], average scores are:[8.97]

*((p+3)[1])=[9.20]  *((p+3)[2])=[9.38]  *((p+3)[3])=[9.40]  *((p+3)[4])=[9.50]  
No.[4] total scores are:[37.480000], average scores are:[9.37]

*((p+4)[1])=[8.89]  *((p+4)[2])=[9.10]  *((p+4)[3])=[9.30]  *((p+4)[4])=[9.40]  
No.[5] total scores are:[36.690002], average scores are:[9.17]

二维数组冒泡排序

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