Nested Vector Logic - C ++

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Nested Vector Logic - C ++

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#1 by (3 votes)
#2 by (0 votes)

4

I have a problem with the following statement:

Given a vector A with n real numbers, get another vector B, also with n real numbers, as follows:

B [1] = 2 * A [1]

B [2] = 3 * A [1] + 2 * A [2]

B [3] = 4 * A [1] + 3 * A [2] + 2 * A [3]

(... and so on)

I made the program, but my logic is wrong and I can not identify the error. Can someone help me? Here is the code I wrote.

#include <iostream>
using namespace std;


int main(){

    int tamanho;

    cout << "Qual o tamanho do vetor?  ";
    cin >> tamanho;

    float vetorA[tamanho], vetorB[tamanho];

    for (int i = 0; i < tamanho; i++){  
        cout<< "Digite o numero :";     
        cin >> vetorA[i];
    }


    for(int i = 0; i < tamanho; i++){
        for(int j = 2; j <= tamanho + 1; j++){
            vetorB[i] += j * vetorA[i];
        }       
    }   


    int i = 0;

    while(i < tamanho){
        cout << "\nA["<< i << "] = " << vetorA[i] << "\t B[" << i << "] = " << vetorB[i]; 
        i++; 
    }
}

c++ array

asked by anonymous 06.09.2015 / 16:35

2 answers

3

Error resolved. I leave the code to help anyone with the same problem.

#include <iostream>
using namespace std;


int main(){

    int tamanho, aux = 0;

    cout << "Qual o tamanho do vetor?  ";
    cin >> tamanho;

    float vetorA[tamanho], vetorB[tamanho];

    for (int i = 0; i < tamanho; i++){  
        cout<< "Digite o numero :";     
        cin >> vetorA[i];
    }


    for(int i = 1; i <= tamanho; i++){
        aux = i + 1;
        for(int j = aux, k = 1; k <= i; j--, k++){
            vetorB[i-1] += j * vetorA[k-1];
        }       
    }   


    int c = 0;
    while(c < tamanho){
        cout << "\nA["<< c << "] = " << vetorA[c] << "\t B[" << c << "] = " << vetorB[c]; 
        c++; 
    }
}

06.09.2015 / 17:59

How to print the entire string if the length exceeds the page Loop Infinite Java

score 0 · Accepted Answer

Perhaps, having thought of a "formula" would help you to get the correct result more quickly.

(n + 1) * A [1] + n * A [2] + (n-1) * A [3] + ... + 3 * A [n-1] 2 * A [n]

The error was really only in logic. By analyzing the statement carefully we can see some patterns.

Graphic explanation http://i62.tinypic.com/2ikf94.jpg

It is noticeable that you have recognized the pattern by looking at your code.

for(int i = 1; i <= tamanho; i++){
    aux = i + 1;
    for(int j = aux, k = 1; k <= i; j--, k++){
        vetorB[i-1] += j * vetorA[k-1];
    }       
}

The for outermost, has as stop condition, the size of the vector. By storing the value of i in an auxiliary variable plus one unit, it covers one of the conditions so that the right result is obtained. In the second for , ensure that the rule will be executed for each expression term until the j and are equal.

Another important way to test your code (as a beginner) to find what is wrong is to crawl.

Good Studies!