8524. Sum of positives in a matrix

A matrix of size n * n is given. Find the sum of all its positive elements.

Input. The first line contains a single integer n (1 ≤ n ≤ 100). The next n lines contain the elements of the n * n matrix. Each element is an integer with an absolute value not exceeding 100.

Output. Print the sum of all positive elements in the matrix.

Sample input	Sample output
3 4 -2 5 1 -4 -12 0 1 -3	11

SOLUTION

array – 2-dim

Algorithm analysis

Using a double loop, compute the sum of the positive elements in the matrix (two-dimensional array).

Algorithm implementation

Declarw a two-dimensional array m.

int m[101][101];

Read the input data. In the variable s, compute the sum of all its positive elements.

scanf("%d",&n);

s = 0;

for(i = 0; i < n; i++)

for(j = 0; j < n; j++)

{

scanf("%d",&m[i][j]);

if (m[i][j] > 0) s = s + m[i][j];

}

Print the answer.

printf("%d\n",s);

Java implementation

import java.util.*;

public class Main

{

public static void main(String[] args)

{

Scanner con = new Scanner(System.in);

int n = con.nextInt();

int m[][] = new int[n][n];

int s = 0;

for(int i = 0; i < n; i++)

for(int j = 0; j < n; j++)

{

m[i][j] = con.nextInt();

if (m[i][j] > 0) s = s + m[i][j];

}

System.out.println(s);

con.close();

}

}

Java implementation – threads

import java.util.*;

class SumRow extends Thread

{

int[][] m;

int row, res;

SumRow(int[][] m, int row)

{

this.m = m;

this.row = row;

}

public void run()

{

res = 0;

for(int i = 0; i < m[0].length; i++)

if (m[row][i] > 0) res = res + m[row][i];

}

public int GetValue()

{

return res;

}

}

public class Main

{

public static void main(String[] args) throws InterruptedException

{

Scanner con = new Scanner(System.in);

int n = con.nextInt();

int m[][] = new int[n][n];

for(int i = 0; i < n; i++)

for(int j = 0; j < n; j++)

m[i][j] = con.nextInt();

SumRow s[] = new SumRow[n];

for(int i = 0; i < n; i++)

{

s[i] = new SumRow(m,i);

s[i].start();

}

for(int i = 0; i < n; i++)

s[i].join();

int sum = 0;

for(int i = 0; i < n; i++)

sum = sum + s[i].GetValue();

System.out.println(sum);

con.close();

}

}

Python implementation

Read the input data.

n = int(input())

a = [list(map(int, input().split())) for _ in range(n)]

In the variable s, compute the sum of all the positive elements in the matrix.

sum = 0

for i in range(n):

for j in range(n):

if a[i][j] > 0: sum += a[i][j]

Print the answer.

print(sum)

Python implementation – list comprehension

Read the input data.

n = int(input())

a = [list(map(int, input().split())) for _ in range(n)]

Compute the sum of all the positive elements in the matrix.

sum = sum(a[i][j] for i in range(n) for j in range(n) if a[i][j] > 0)

Print the answer.

print(sum)