Created
February 18, 2025 16:58
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import java.io.*; | |
import java.util.*; | |
import java.util.Arrays; | |
public class Solution { | |
public static void main(String[] args) { | |
//1. Formar un arreglo principal | |
Scanner scanner = new Scanner(System.in); | |
int n = scanner.nextInt(); | |
int numbers [] = new int [n]; | |
for ( int i = 0; i < n; i++){ | |
numbers[i] = scanner.nextInt(); | |
} | |
int sum; | |
int count = 0; | |
//2. Extraer subarreglos | |
for (int i=0; i <= n; i++) | |
{ | |
// Pick ending point | |
for (int j=i; j<=n; j++) | |
{ | |
sum = 0; | |
for (int num : Arrays.copyOfRange(numbers, i, j)) { | |
sum += num; | |
} | |
if (sum < 0) { | |
count++; | |
} | |
// System.out.println(sum); | |
} | |
} | |
//3. Devolver el numero posible de subsrrays con suma negativa | |
System.out.println(count); | |
} | |
} |
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