CODEWITHSUNDEEP
c++
Faraz Mazhar
Faraz Mazhar

Reputation: 89

Add ten numbers using only two variables and a loop

So, I used following code but my friends say this code is incorrect as I am using third variable in for-loop. Kindly help.

int x = 0;
int y = 0;
for ( int i = 0; i < 10; i++) 
{
  cin >> y;
  x = x + y;
}
cout << x;

Upvotes: 0

Views: 1178

Answers (5)

Frerich Raabe
Frerich Raabe

Reputation: 94329

I see no rules restricting the types of variables, so you could just use an array and do it in one variable:

int arr[3] = {0};
for (; arr[0] < 10; ++arr) {
  cin >> arr[1];
  arr[2] += arr[1];
}
cout << arr[2];

However, you could also do it using a recursive function using two variables (if you consider the function argument a variable); this doesn't even need a loop (though I'm not sure if the and a loop part means that you have to use a loop, in which case you could of course just put a sneaky do { ... } while(0); somewhere. ;-)

int sum(int x) {
    if (x == 0) {
        return 0;
    }

    int v;
    cin >> v;
    return v + sum(x - 1);
}

cout << sum(10);

Upvotes: 0

Khalil Khalaf
Khalil Khalaf

Reputation: 9407

If I understand correct you want to use this x += y; . This is a two-variable inside a loop method that does the same functionality as x = x + y; that they considered a three-variables inside a loop.

Upvotes: 0

AnT stands with Russia
AnT stands with Russia

Reputation: 320531

If you can restrict the range of the sum, you can do something like

const unsigned LIMIT = 1000000;

unsigned x;
for ( x = LIMIT * 10; x >= LIMIT; x -= LIMIT) 
{
  unsigned y;
  cin >> y;
  x += y;
}
cout << x;

This would be an arithmetic equivalent of other bit-fiddling tricks based on squeezing two values into one variable, as in @Richard Hodges's answer.

But that requires restricting the range of the sum.

Upvotes: 1

Richard Hodges
Richard Hodges

Reputation: 69882

Rationale:

Use a 64-bit value as the accumulator since it's unlikely that users will enter numbers that large.

Use the top 8 bits (could be 4) of the accumulator as the counter.

#include <iostream>

int main()
{
    using namespace std;

    int64_t accum = int64_t { 10 } << 56;
    while (accum & (int64_t{0xff} << 56))
    {
        int next;
        cin >> next;
        accum += next;
        accum -= int64_t { 1 } << 56;
    }

    cout << accum << endl;


    return 0;
}

Upvotes: 1

Russley Shaw
Russley Shaw

Reputation: 441

I see nothing wrong with this code. More descriptive variables would be more helpful maybe (i.e. y -> userInput, x -> sum).

Upvotes: 2

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