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Asked: 2020-01-23 09:25:02 +0800 CST 2020-01-23 09:25:02 +0800 CST

How do you do this little program?

  • 772

It is an exercise that I cannot solve, besides that my knowledge about matrices is still zero and I don't know where to start.

public class ControlMatriz {
    @SuppressWarnings({ "null" })
    public void cien1NumNatural(int x, int supx)
    {
        for(int i=0;i<x;i++){
                for(int j=0;j<supx;j++){
                    int[][] matrizM = null;
                    System.out.println(matrizM[i][j]);
                    if (matrizM[i][j] % 2==0)
                        System.out.println(matrizM[i][j]*5);
                    }
                 }
              }
    }
java

1 Answers

  1. Best Answer

    First: Calculate how many? and which one is it? will be the bases.

    int total = num_introducido / 2;    // calcula cuantas pares hay
int[] bases = new int[total];       // crea el arreglo
for(int i = 1; i <= total; i ++)
    bases[i-1] = i * 2;               // carga el arreglo

    Second: Calculate the powers.

    int[][] potencias = new int[total][5];

for(int i = 0; i < total; i++){ 
      potencias[i][0] = bases[i];
      potencias[i][1] = (int) Math.pow(bases[i], 2);
      potencias[i][2] = (int) Math.pow(bases[i], 3);
      potencias[i][3] = (int) Math.pow(bases[i], 4);
      potencias[i][4] = (int) Math.pow(bases[i], 5);
}

    Third: You print the information.

    for(int i = 0; i < total; i++){

      System.out.print(potencias[i][0] + " - ");
      System.out.print(potencias[i][1] + " - ");
      System.out.print(potencias[i][2] + " - ");
      System.out.print(potencias[i][3] + " - ");
      System.out.println(potencias[i][4]);

}

    Although you don't need the base array at all, in the second loop, you could replace it with:

    ((i + 1) * 2)

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