关于计算数组大小的C ++问题 [英] C++ question about counting array size

问题描述

大家好，我犯了逻辑错误。对我的朋友来说这是一个简单的项目，但我找不到为什么我做错了。也许你可以帮助我。

所以我有矩阵，我想计算我的诊断以下所有矩阵负数的算术平均值。所以我正在做的是得到所有低于诊断数字并总结它们，latar我得到它的数量，只是div sum / countn。但是我不仅计入了负数，而且还得到了积极的数据：

1 2 3

-4 5 6

2 -4 6



以下的诊断; -4 2 -4



所以负数是2，但是我的代码是3（带负数和正数的3）。





提前感谢您的帮助:)



我的尝试：

Hello guys I am getting logical mistake. It is easy project for my friend, but I can't find why I am doing it wrong. Maybe you can help me.
So I have matrix and I want to count arithmetical average of all matrix negative number below my diagnol. So What I am doing is getting all below diagnol numbers and sum them, latar I am getting its count and just div sum / countn. But I am getting in my count not only negative numbers, but also an positives for ex:
1 2 3
-4 5 6
2 -4 6

below diagnol; -4 2 -4

so negative count is 2, but my code says it is 3 (it is 3 with negatives and positives).


Thanks for help in advance :)

What I have tried:

for (int i =0 ; i<row ; i++)
{
    for (int j=0 ; j<col ; j++)
{
    if ( i > j )
    {
        if (M[i][j]<0)
            sum= sum + M[i][j];
            countn++;
    }
    }
}
 cout<<"neg suma: "<< sum <<endl<<"neg number: "<< countn<< endl << "Average for Negative: " << sum / countn << endl;
return 0;
}

推荐答案

替换

replace

引用：

if（M [i] [j]< 0）

sum = sum + M [i] [j];

countn ++;

if (M[i][j]<0)
sum= sum + M[i][j];
countn++;

with

with

if (M[i][j]<0)
{
  sum= sum + M[i][j];
  countn++;
}

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