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

A. From Hero to Zero

题目链接：http://codeforces.com/contest/1175/problem/A

题目

You 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

题意

. 给你两个数n,k,你需要将n每次经过以下两个步骤之一从而得到0，输出变换次数：
要么n=n-1，要么将n=n/k(前提n能整除k）。

思路

范围太大，暴力绝对TLE，尝试就是浪费生命！
巧办法：
n%k！=0时，变换的步数就是n%k的值，此时当前n=（n-减掉该步数）
n%k==0时，变换的步数就是1,此时当前n=n/k，
n为0结束，将步数累加输出即可。

#include
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;
}