求其平均值。一个int数组的最大,最小的数字和模式 [英] Find the average. largest, smallest number and mode of a int array
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问题描述
我试图让一个程序,用户最多可以输入25个数字,然后告诉用户平均,最小和最大数量和输入的数字的模式,但我不知道如何做到这一点
任何指导,将不胜感激。
这里是我迄今为止
静态无效的主要(字串[] args)
{
Console.WriteLine(输入数字的金额,你想用的:);
INT arraylength = Int32.Parse(到Console.ReadLine());
INT [] = AverageArray新INT [25];
//填充用户输入
的阵列(INT I = 0; I< AverageArray.Length;我++)
{
Console.Write(输入要查找的平均水平的数字:);
AverageArray [I] = Int32.Parse(到Console.ReadLine());
}
//打印出数组
Console.WriteLine(这里是平均:);
的for(int i = 0; I< AverageArray.Length;我++)
{
Console.WriteLine(AverageArray [I]);
}
Console.WriteLine(FindAverage(AverageArray));
}
公共静态双FindAverage(INT [] averageNumbers)
{
INT arraySum = 0;
的for(int i = 0; I< averageNumbers.Length;我++)
arraySum + = averageNumbers [I]
返回arraySum / averageNumbers.Length;
}
公共静态双LargestNumber(INT []订购数量,诠释计数)
{
}
公共静态双SmallestNumber(INT []订购数量,诠释计数)
{
}
公共静态双模式(INT []订购数量,诠释计数)
{
}
}
解决方案
公共静态双LargestNumber(INT []订购数量,诠释计数)
{
双上限=订购数量[0];
的for(int i = 1; I<计数;我++)
如果(订购数量[I]>最大)最大=订购数量[I]
返回分钟;
}
公共静态双SmallestNumber(INT []订购数量,诠释计数)
{
双分钟=订购数量[0];
的for(int i = 1; I<计数;我++)
如果(订购数量[1] - ;分)分=订购数量[I]
返回分钟;
}
公共静态双模式(INT []订购数量,诠释计数)
{
双模式=订购数量[0];
INT MAXCOUNT = 1;
的for(int i = 0; I<计数;我++)
{
INT VAL =订购数量[I]
诠释计数= 1;
为(INT J = I + 1; J<计数; J ++)
{
如果(订购数量[J] == VAL)
计数++;
}
如果(计数> MAXCOUNT)
{
模式= VAL;
MAXCOUNT =计数;
}
}
返回模式;
}
Hi I am trying to make a program where the user can enter up to 25 numbers and then it tells the user the average, smallest and largest number and the mode of the numbers entered but I am not sure on how to do this.
Any Guidance would be appreciated
here is what I have so far
static void Main(string[] args)
{
Console.WriteLine("enter the amount of numbers you would like to use of: ");
int arraylength = Int32.Parse(Console.ReadLine());
int[] AverageArray = new int[25];
//filling the array with user input
for (int i = 0; i < AverageArray.Length; i++)
{
Console.Write("enter the numbers you wish to find the average for: ");
AverageArray[i] = Int32.Parse(Console.ReadLine());
}
//printing out the array
Console.WriteLine("here is the average: ");
for (int i = 0; i < AverageArray.Length; i++)
{
Console.WriteLine(AverageArray[i]);
}
Console.WriteLine(FindAverage(AverageArray));
}
public static double FindAverage(int[] averageNumbers)
{
int arraySum = 0;
for (int i = 0; i < averageNumbers.Length; i++)
arraySum += averageNumbers[i];
return arraySum / averageNumbers.Length;
}
public static double LargestNumber(int[] Nums, int Count)
{
}
public static double SmallestNumber(int[] Nums, int Count)
{
}
public static double Mode(int[] Nums, int Count)
{
}
}
解决方案
public static double LargestNumber(int[] Nums, int Count)
{
double max = Nums[0];
for (int i = 1; i < Count; i++)
if (Nums[i] > max) max = Nums[i];
return min;
}
public static double SmallestNumber(int[] Nums, int Count)
{
double min = Nums[0];
for (int i = 1; i < Count; i++)
if (Nums[i] < min) min = Nums[i];
return min;
}
public static double Mode(int[] Nums, int Count)
{
double mode = Nums[0];
int maxCount = 1;
for (int i = 0; i < Count; i++)
{
int val = Nums[i];
int count = 1;
for (int j = i+1; j < Count; j++)
{
if (Nums[j] == val)
count++;
}
if (count > maxCount)
{
mode = val;
maxCount = count;
}
}
return mode;
}
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