2 For loops one result - c

Question

I need your wisdom. Looks not a big problem but I need a way.
First, I will share code. This code is correct but I need some addition, inside for loops have criteria if voltages are bigger than percentage it's ok but all be correct I need just one writing. I have 2 loops but need just one prompt.
If it is confusing I can share original question. Thank you guys.
I put original question:
Voltage readings are obtained from an electrical substation once every hour for six hours (so there are six
readings). Write a C program to perform the following checks on the substation:
a) display all voltages that differ from the average by more than 10% of the average.
b) display all pairs of consecutive hours where the change from the voltage at one hour
to the next is greater than 15% of the average.
Example 1
Enter 6 voltages: 210.1 223.2 189.6 206.2 235.1 215.0
The average is 213.2 volts.
10% = 21.3 volts.
15% = 32.0 volts.
The following problems occurred:
1. Voltage at hour 3 was 189.6 volts (difference of 23.6 volts).
2. Voltage at hour 5 was 235.1 volts (difference of 21.9 volts).
3. Voltage change from hour 2 to hour 3 was 33.6 volts.
Example 2
Enter 6 voltages: 233.1 201.0 221.5 240.2 222.7 208.1
The average is 221.1 volts.
10% = 22.1 volts.
15% = 33.2 volts.
No problems were encountered.
#include <stdio.h>
#include <math.h>
#include <string.h>
int i;
float volt[6];
float avg, avg10, avg15, total, a, b;
int main () {
total= 0 ;
avg = 0;
printf("Enter 6 Volts of Machine\n");
for ( i=0; i<6; i++) {
printf("Type %d. volt", i+1);
scanf("%f",&volt[i]);
total = total + volt[i];
}
avg = total/6;
avg10 = (avg * 10) / 100;
avg15 = (avg * 15) / 100;
printf("------------------------------------------\n");
printf("The machine Avarage Voltage is %.2f\n", avg);
printf("The Machine Avarage is%.2f\n", avg10);
printf("The Machine 15 Avarage is%.2f\n\n\n", avg15);
for (i=0;i<6;i++) {
a = fabs(volt[i] - avg);
if( a > avg10 ) {
printf("\nVoltage at hour %d was %.2f volts (diffrence of %.2f volts)\n\n", i+1, volt[i], a);
}
}
for (i=0; i<5; i++) {
b = fabs(volt[i+1] - volt[i]);
if( b > avg15) {
printf("\nVoltage change from hour %d to hour %d was %.2f\n\n", i+1, i+2, b);
}
}

Answer 1

If you need just one loop try something like this:
for (i=0;i<6;i++)
{
if((a = fabs(volt[i] - avg)) > avg10 )
{
printf("\nVoltage at hour %d was %.2f volts (diffrence of %.2f volts)\n\n", i+1, volt[i], a);
}
if((i < 5 && (b = fabs(volt[i+1] - volt[i])) > avg15 )
{
printf("\nVoltage change from hour %d to hour %d was %.2f\n\n", i, i+1, b);
}
}

Answer 2

If you want to print out a message when no problems were encountered, you must remember if any or how many errors were reported. You cannot print out such messages inside the loop, of course, because saying "No errors occurred" eight times and reporting thre errors is a bit of a contradiction.
Your expected out put shows an enumeration of errors, so it is a good idea to keep a count of errors. Proceed as follows:
Whenever you print an error message, increase the count of errors.
Before you do so, check whether this is the frst error that is reported. If so, print the caption ("The following errors occurred")
If you have checked everything and no errors have occurred, print the success message.
Or, in code:
int nerror = 0;
for (i = 0; i < n; i++) {
double v = fabs(volt[i] - avg);
if (v > avg10) {
if (nerror == 0) {
puts("The following problems occurred:");
}
nerror++;
printf("%d. Voltage at hour %d was %.2f volts "
"(diffrence of %.2f volts)\n",
nerror, i + 1, volt[i], v);
}
}
for (i = 1; i < n; i++) {
double diff = fabs(volt[i - 1] - volt[i]);
if (diff > avg15) {
if (nerror == 0) {
puts("The following problems occurred:");
}
nerror++;
printf("%d. Voltage change from hour %d to "
"hour %d was %.2f\n",
nerror, i, i + 1, diff);
}
}
if (nerror == 0) puts("No problems were encountered.");

Answer 3

Thank you for everyone my question was solved. Happy coding!
Code is :
#include <stdio.h>
#include <math.h>
int i;
float volt[6];
float avg, avg10, avg15, total, a, b;
int main () {
int voltageproblem1 = 0;
int voltageproblem2 = 0;
total= 0 ;
avg = 0;
printf("Enter 6 Volts of Machine\n");
for ( i=0; i<6; i++) {
printf("Type %d. volt", i+1);
scanf("%f",&volt[i]);
total = total + volt[i];
}
avg = total/6;
avg10 = (avg * 10) / 100;
avg15 = (avg * 15) / 100;
printf("------------------------------------------\n");
printf("The machine Avarage Voltage is %.1f\n", avg);
printf("The Machine Avarage is%.1f\n", avg10);
printf("The Machine 15 Avarage is%.1f\n\n\n", avg15);
for (i=0;i<6;i++) {
a = fabs(volt[i] - avg);
if( a > avg10 ) {
printf("\nVoltage at hour %d was %.1f volts (diffrence of %.1f volts)\n\n", i+1, volt[i], a);
voltageproblem1 =1;
}
}
for (i=0; i<5; i++) {
b = fabs(volt[i+1] - volt[i]);
if( b > avg15) {
printf("\nVoltage change from hour %d to hour %d was %.1f\n\n", i+1, i+2, b);
voltageproblem2 = 1;
}
}
if ((voltageproblem1==0)&&(voltageproblem2==0)) {
printf("No problems were encountered.\n\n");
}
}