Finding Sum

Question

Finding Sum
Problem Description
You are given a set of N positive integers and another integer P, where P is a small prime. You need to write a program to count the number of subsets of size (cardinality) 3, the sum of whose elements is divisible by P. Since the number K of such sets can be huge, output K modulo 10007 1009 (the remainder when K is divided by 1009)

Constraints
N <= 500

P<50

All integers <=1000

Input Format
First line two comma separated integers N, P

The next line contains N comma separated integers

Output
One integer giving the number of subsets the sum of whose elements is divisible by P. Give result modulo 1009

Explanation
Example 1

Input

4,5

5,10,15,20

Output

4

Explanation

Every non empty subset of the given numbers has sum of its elements a multiple of 5. Since there are 4 subsets of size 3, the output is 4.

Example 2

Input

5,5

3,7,12,13,15

Output

4

Explanation

There are 4 subsets of size 3 with sum a multiple of 5: {3, 7, 15}, {12, 13, 15}, {7, 13, 15}, {3, 12, 15}, Hence the output is 4.

Answer 1

only for some test cases

#include<stdio.h>
int main(){
int n,i,j,k,sum=0,a[100],p,c=0;
scanf("%d,%d\n",&n,&p);
for(i=0;i<n;i++)
{
scanf("%d,",&a[i]);
}
for (i = 0; i < n; i++) {
for (j = i + 1; j < n;) {
if (a[j] == a[i]) {
for (k = j; k < n; k++) {
a[k] = a[k + 1];
}
n--;
} else
j++;
}
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
for(k=j+1;k<n;k++)
{
sum=a[i]+a[j]+a[k];
if(sum%p==0)
c++;
sum=0;
}
}
}
printf("%d",c);
return 0;
}

Answer 2

import java.util.Scanner;

public class MyClass {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String input = sc.next();
String ans[] = input.split(",");
int n = Integer.parseInt(ans[0]);
int p = Integer.parseInt(ans[1]);
int arr[] = new int[n];
String arrin = sc.next();
String an[] = arrin.split(",");
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(an[i]);
if(arr[i]>1000)
return;
}
int cnt = 0;
if(n>500 || p>49)
return;
else{
for (int i = 0; i < arr.length - 2; i++) {
for (int j = 1; j < arr.length - 1; j++) {
for (int k = 2; k < arr.length; k++) {
if ((arr[i] + arr[j] + arr[k]) % p == 0 && (i != j) && (j != k) && (i != k)) {
cnt++;
//System.out.println(arr[i] + " " + arr[j] + " " + arr[k]);
}
}
}
}
System.out.println(cnt%1009);
}
}

}

Answer 3

#include<stdio.h>
int main(){
int n,i,j,k,sum=0,a[100],p,c=0;
scanf("%d,%d\n",&n,&p);
for(i=0;i<n;i++)
{
scanf("%d,",&a[i]);
}
for (i = 0; i < n; i++) {
for (j = i + 1; j < n;) {
if (a[j] == a[i]) {
for (k = j; k < n; k++) {
a[k] = a[k + 1];
}
n--;
} else
j++;
}
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
for(k=j+1;k<n;k++)
{
sum=a[i]+a[j]+a[k];
if(sum%p==0)
c++;
sum=0;
}
}
}
printf("%d",c);
return 0;
}

Answer 4

import java.util.Scanner;
public class MyClass {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String input = sc.next();
String ans[] = input.split(",");
int n = Integer.parseInt(ans[0]);
int p = Integer.parseInt(ans[1]);
int arr[] = new int[n];
String arrin = sc.next();
String an[] = arrin.split(",");
if(an.length != n)
return;
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(an[i]);
if(arr[i]>1000)
return;
}
int cnt = 0;
boolean pIsprime = true;
int c=2;
while(pIsprime && c<=p/2){
if(p%c == 0)
pIsprime = false;
c++;
}
if(n>500 || p>49 || n<0 || !pIsprime || p<2)
return;
else{
for (int i = 0; i < arr.length - 2; i++) {
for (int j = 1; j < arr.length - 1; j++) {
for (int k = 2; k < arr.length; k++) {
if ((arr[i] + arr[j] + arr[k]) % p == 0 && (i != j) && (j != k) && (i != k)) {
cnt++;
//System.out.println(arr[i] + " " + arr[j] + " " + arr[k]);
}
}
}
}
System.out.println(cnt%1009);
}
}

}

Answer 5

It works can you show me the error.

Answer 6

This code not working
Help me to solve

Answer 7

Write a C program to sum first 100, 500 and then 1000 terms of sum
Sk
N=∑
1
N
1/i
k
. It should take k and N as input and give the sum as output. Plot the output
with respect to i=1:N. For k=2, draw a horizontal line at y=(∏^2)/6 . Do similar plots for k=4,
6, 8. 10.