Calculate two-dimensional array smallest elements in each column, and in turn into a one-dimensional array p,

Subject content:

Write function void fun (int t [M] [N], int p [N]), line point t a M N columns of two-dimensional arrays, and the two dimensional array smallest elements in each column, and in turn into a one-dimensional array p, two-dimensional array of data given in the main function, and each column of the smallest element in the output of the main function, define symbolic constant M is 5, symbolic constant N is 6,

CodePudding user response:

Conveniently wrote a, consult

 5 # define M 
# define N 6 
Void fun (int t [M] [N], int p [N]) 
{
Int r, c, m; 
For (c=0; C{
M=t [0] [c]; 
For (r=1; r{
If (t [r] [c] {
M=t [r] [c]; 
} 
} 
P=m [c]; 
} 
} 

Int main () 
{
Int r, c, t [M] [N]={0}, p={0} [N]. 
For (r=0; r{
For (c=0; C{
The scanf (" % d ", & amp; T [r] [c]); 
} 
} 
Fun (t, p); 
For (c=0; CPrintf (" % d ", p [c]); 

return 0; 
} 

CodePudding user response:

A one-dimensional array to use the first line of the two-dimensional array populated, and fixed n a good

 
# include & lt; stdio.h> 
# define M 5 
# define N 6 
Void fun (int t [M] [N], int p [N]) {
Int I=0, j=0; 
for(i=0; IP=t [0] [I] [I]; 
}//assignment, if you will be a pointer can point to the first line directly 
for(j=0; Jfor(i=0; IIf (p [j] & gt; [I] [j])=t 
{
P=t [j] [I] [j]; 
} 
} 
} 
} 
Int main () {
int i=0; Int j=0; 
Int t [M] [N]={}; 
Int [N] p={}; 
for(i=0; Ifor(j=0; JThe scanf (" % d ", & amp; T [I] [j]); 
} 
} 
Fun (t, p); 
for(i=0; IPrintf (" % d ", p [I]); 
} 
printf("\n"); 
return 0; 
} 

Upstairs directly define m is also a principle, that is, to the back of the comparison