Input a number of natural Numbers, the last is 0 (end of said input), the output is greater than the-CodePudding

#include
using namespace std;
Int main ()
{
Int a, b;
Float c;
int n;
For (a=0, b=0;; +, b +=n, c=1.0 *
b/a){
Cin> n;
If (n==0) break;
}
Cout<& lt;" The average="& lt; return 0;
}
This want how to just can make the output is the result of the "greater than the average number of"

CodePudding user response:

Using a container to the input values are stored, calculate the average, in comparing the

CodePudding user response:

Or an array input number

CodePudding user response:

Not very understand, can you help me to change

CodePudding user response:

For reference array version of

 
#include  
using namespace std; 
Const int N=1024; 

Int main () 
{
Int arr [N]. 
Int I, a=0, b=0, count=0; 
Float c; 

for (i=0; I{
Cin & gt;> Arr [I]; 
B +=arr [I]; 

If (arr [I]==0) 
{
A=I; 
break; 
} 
} 

C=b * 1.0/a; 
for (i=0; I{
If (arr [I] & gt; C) 
count++; 
} 

Cout<& lt;" The average="& lt; Cout & lt; <"Greater than the average number of:" & lt; return 0; 
}

CodePudding user response:

reference 4 floor CHXCHXKKK response:

array for reference version of the

 
#include  
using namespace std; 
Const int N=1024; 

Int main () 
{
Int arr [N]. 
Int I, a=0, b=0, count=0; 
Float c; 

for (i=0; I{
Cin & gt;> Arr [I]; 
B +=arr [I]; 

If (arr [I]==0) 
{
A=I; 
break; 
} 
} 

C=b * 1.0/a; 
for (i=0; I{
If (arr [I] & gt; C) 
count++; 
} 

Cout<& lt;" The average="& lt; Cout & lt; <"Greater than the average number of:" & lt; return 0; 
}

It is recommended to use the vector

 
#include 
#include 
#include 
using namespace std; 
Int main () 
{
vector a; 
int n; 
While (cin> N) 
Amy polumbo ush_back (n); 
Float the count; 
The count=the accumulate (a.c the begin (), a.c end (), 0.0) * 1.0/a.s considering (); 
Int num. 
For (const auto& P: a) 
If (p> The count) 
+ + num. 
Cout<& lt;" The mean: "& lt; Cout<& lt;" Is greater than the average number of: "& lt; return 0; 
}