How do I output the minimum value of each column in the m m square matrix with C? 5

The solution is as follows:

#include using namespace std;int main() for (i=0;i<3;i++)

cout<<"As a result, it was"<

#include

#include

#define m 5

int fun(int n,int xx[m])for(i=0;isum+=xx[i][i]+xx[i][n-i-1];

return(sum);

void main(),system("cls");

printf("the sum of all elements on 2 diagnals is %d",fun(m,aa));

The situation is as follows:

the5x5 matrix:

The sum of all elements on 2 diagnals is 50 press any key to continue. .

int i,j;

int a[5][5]=,,, is a random assignment here.

for(i=0;i<5;i++)

for(j=0;j<5;j++)

cout

min(if(a1:a1000=10,c1:c1000,9e+307))

Array formula, ctrl+shift+enter to end the input.

#include

#include

using namespace std;

int rand2(int m)

int main()

for (int j = 0;j < 10;j++)int max =min[0];

int min =max[0];

for (int i = 1;i < 10;i++)if (max[i] min = max[i];

cout <<"The smallest of the 10 maximums is:"<< min max)

max = min[j];

cout <<"The maximum of the 10 minimum values is:"<< max

cout <<"The square of the difference between two numbers:"

return 0;

This question is actually very simple!

It's just to assign a value, then find out the size and compare!

I'm at work now, and I'll help you write it out after work!

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Not for the score, just for the problem!

Thank you and have a great day!

Boring question. Who wants to write such a whole bunch of boring **.

There are two formulas to keep in mind.

It is a special Laplace style.

Just remember.

The preceding is the form of the determinant, not the form of the matrix.

c=zeros(1,2);

c(2)=min(b(:,2));

c(1)=find(b(:,2)==c(2),1);

I'm only outputting the number of rows with the first minimum value from top to bottom).

Eros of the Arrow Ha Shi Da Cassadine General.

Because it's a quadratic function and the coefficient before m squared is greater than 0, the function has a minimum value at -b 2a, and the minimum value is 2, where m = 3