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# Write a program to get N number of positive non-zero integers, so that the sum and product of these numbers are equal

## Write a program to get N number of positive non-zero integers, so that the sum and product of these numbers are equal Tag : php , By : user121350 Date : November 23 2020, 09:01 AM

The answer is number of numbers, 2 and (number of numbers - 2) of 1s!
``````<?php

\$n = 5;

\$numbers = array();
\$numbers[] = \$n;
\$numbers[] = 2;
for(\$i=1; \$i<=(\$n-2); \$i++){
\$numbers[] = 1;
}
``````

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## Write a program to find the sum of positive odd numbers and the product of positive even numbers less than or equal to 3

Tag : c , By : afarouk
Date : March 29 2020, 07:55 AM
hop of those help? Try using %ld instead of %d in your printf:
printf("\nThe product of positive even numbers is: %ld", product);

## Write a program that keeps taking in integers until a negative integer is given. Then take all the positive integers and

Tag : java , By : Sandip
Date : March 29 2020, 07:55 AM
I hope this helps you . You need to add number = sc.nextInt(); inside the while loop. Otherwise it would be an infinite loop.
Another point you have missed is integer division. Most of the above answers have missed it too. Both your counter variable and sum variable are integers. Therefore the division will also be an integer division.
``````sum = 17;
counter = 4;
``````
``````int average = 0;
average = (double)sum/counter;
``````

## Write a program that multiplies user entered number till product of these numbers reach 1000

Tag : c , By : John Miller
Date : March 29 2020, 07:55 AM

## Write a function that input a positive integer n and return the number of n-digit positive integers that divisible by 17

Tag : python , By : vbanos
Date : March 29 2020, 07:55 AM

## Write a C program that read 5 numbers and counts the number of positive numbers and print the average of all positive va

Tag : c , By : Mariamario
Date : October 01 2020, 06:00 AM
like below fixes the issue While there is nothing wrong with repetitive lines of code, it's certainly not literature. If you are faced with doing the same thing 5 times and the only thing changing is an index or two in the output -- you should be thinking loop. Here you can expand the loop you have to include the prompting for input as well as handling the summing of positive numbers, and it can all be done without having to store the input values in an array.
All you need to maintain within the loop is the sum of the positive values. The after you exit the loop you can compute the average based on the number of positive values input.
``````#include <stdio.h>

#define NNUM 5      /* if you need a constant, #define one (or more) */

int main (void) {

int n = 0,      /* no. of positive value counter */
sum = 0,    /* sum of all positive values */
val;        /* each input value */
double avg;     /* floating-point number to hold average */

for (int i = 0; i < NNUM; i++) {    /* loop NNUM times */
printf ("enter no. %d: ", i+1); /* prompt for input */
if (scanf ("%d", &val) != 1) {  /* validate EVERY user input */
fputs ("error: invalid integer input.\n", stderr);
return 1;
}
if (val > 0) {      /* check if val is positive */
sum += val;     /* add val to sum */
n++;            /* increment pos value count */
}
}
avg = (double)sum / n;  /* compute average (note cast) */

printf ("\nThe number of positive numbers       : %d\n"
"The average of all positive value is : %.2f\n", n, avg);
}
``````
``````\$ ./bin/avgposnum
enter no. 1: 14
enter no. 2: 63
enter no. 3: 78
enter no. 4: 45
enter no. 5: 21

The number of positive numbers       : 5
The average of all positive value is : 44.20
``````
``````\$ ./bin/avgposnum
enter no. 1: -10
enter no. 2: 10
enter no. 3: 0
enter no. 4: 20
enter no. 5: -20

The number of positive numbers       : 2
The average of all positive value is : 15.00
``````