piyushgoyal443's blog

By piyushgoyal443, history, 5 years ago, In English

this is my solution and it works


#include < iostream >

using namespace std;

int main()
{
    int a, b, div, rem;
    cin >> a >> b;
    int sum = a;
    while(a >= b)
    {
        div = a / b;
        sum += div;
        rem = a % b;
        a = div + rem;
    }
    cout << sum << endl;
    return 0;
}

[link to the problem] (http://codeforces.com/problemset/problem/379/A)
`

but I saw other people solutions and they have used a formula that calculates answer without any loop

 ans = (a * b — 1)/(b — 1) 

Can someone please help me how to get this formula?

 
 
 
 
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5 years ago, # |
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Auto comment: topic has been updated by piyushgoyal443 (previous revision, new revision, compare).

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Please provide a link to the problem.

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There is no var called 'mod' in your code

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5 years ago, # |
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(sorry for my english) Given:

a = q1 * b + r1;   (1)
q1 + r1 = q2 * b + r2;   (2)
       ...
q(n - 1) + r(n - 1) = qn * b + rn;   (n) where qn = 0

ans = a + sum(i = 1 ... n - 1){qi}

Let s = sum(i = 1 ... n — 1){qi}

So we need to find s. From (1) we get r1:

r1 = a - q1 * b

From this and (2) we get r2:

r2 = a - q1 * (b - 1) - q2 * b

And so we get r3:

r3 = a - (q1 + q2) * (b - 1) - q3 * b

And so on:

rn = a - s * (b - 1) - qn * b = a - s * (b - 1)

Then

s = (a - rn) / (b - 1) (s)

if rn equals 0 then q(n — 1) + r(n — 1) = 0 too. And then q(n — 1) = 0 and n = n — 1 what is wrong. Then

1 <= rn < b

Then we need to subtract at least 1 from a in (3):

s = (a - 1) / (b - 1)

answer = a + s = a + (a - 1) / (b - 1) = (a * b - 1) / (b - 1)