Im trying to figure out how to get the program to tell me ho

Im trying to figure out how to get the program to tell me how many even and odd numbers were entered by the user. Any ideas or tips would help.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#include <stdio.h>
int main ()
{
int n, integer, sumOdd=0, sumEven=0;
//asks the users name
char name [100];
printf(\"What is your name? \");
scanf(\"%s\", name);
printf(\"Please enter 0 to exit at anytime.\ \");
printf(\"Enter any integer: \");
scanf(\"%d\", &integer);
while(integer != 0)
{
if(integer %2 == 0)
{
sumEven = sumEven + integer;
}
if(integer %2 != 0) /*You need not perform this check, you can simply write an else.*/
{
sumOdd = sumOdd + integer;
}
//asks users for second number
printf(\"Enter any integer: \");
scanf(\"%d\",&integer);
}
printf(\"%s, The numbers you entered are broken down as follows:\ You entered even numbers with a total value of %i.\ You also entered odd numbers with a total value of %i. \ \",name,sumEven, sumOdd);

return 0;
}

Solution

Count how many even and odd numbers:
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
int main ()
{
int n, integer, sumOdd=0, sumEven=0,evenCount=1,oddCount=1;
//asks the users name
char name [100];
printf(\"What is your name? \");
scanf(\"%s\", name);
printf(\"Please enter 0 to exit at anytime.\ \");
printf(\"Enter any integer: \");
scanf(\"%d\", &integer);
for(int i=1;i<=integer;i++)
{
if(integer %2 == 0)
{
sumEven = sumEven + integer;
evenCount++;

}
else if(integer %2 != 0) /*You need not perform this check, you can simply write an else.*/
{
sumOdd = sumOdd + integer;
oddCount++;

}
}

printf(\"count the number of EvenNumbers\"+evenCount);
printf(\"sum of the number of EvenNumbers\"+sumEven);
printf(\"count the number of oddNumbers\"+oddCount);
printf(\"sum of the number of oddNumbers\"+sumOdd);
}