Educational Codeforces Round 66 (Rated for Div. 2) A

A. From Hero to Zero

Topic links: http://codeforces.com/contest/1175/problem/A

topic

ou are given an integer n and an integer k
In one step you can do one of the following moves:
decrease n by 1;
divide n by k if n is divisible by k.
For example, if n=27 and k=3 you can do the following steps: 27→26→25→24→8→7→6→2→1→0.
You are asked to calculate the minimum number of steps to reach 0 from n.

input

The first line contains one integer t (1≤t≤100) — the number of queries.
The only line of each query contains two integers n
and k (1≤n≤1018, 2≤k≤1018).

output

For each query print the minimum number of steps to reach 0
from n in single line

Example

intput

2
59 3
1000000000000000000 10


output

8

19

The meaning of problems

Give you two numbers n, k, n you need to go through each one of the following two steps ** ** 0 whereby the output frequency converted:
either n = n-1, either the n = n / k (premise n divisible k).

Thinking

Too large, violent absolute TLE, try is a waste of life!
Clever way:
the n-% k! = 0, the number of steps is the conversion of n% k value, the current case n = (n- minus the number of steps)
n% k == 0, the conversion is a number of steps, the current case n = n / K,
n-0 is completed, the accumulated number of steps can be output.

 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+7;
int main()
{
    int T;
    cin>>T;
    while(T--) {
        ll n, k;
        cin >> n >> k;
        ll result = 0;
        while (n != 0) {
            ll book = n % k;
            if (book != 0) {
                result += book;
                n = n - book;
            } else {
                result += 1;
                n = n / k;
            }
        }
        cout << result << endl;
    }
return 0;
}

 

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Origin www.cnblogs.com/Vampire6/p/10990374.html