Codeforces Round #466 (Div. 2) B. Our Tanya is Crying Out Loud_right now she actually isn't. but she will be, if -优快云博客

本文链接：https://blog.youkuaiyun.com/s540239976/article/details/79367866

本文介绍了一个算法问题，目标是最小化将一个整数n通过特定操作转换为1所需的代价。两种操作包括减1和整除，每种操作都有不同的成本。文章提供了完整的代码实现，并解释了如何采用贪心策略来解决这个问题。

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Right now she actually isn't. But she will be, if you don't solve this problem.

You are given integers n, k, A and B. There is a number x, which is initially equal to n. You are allowed to perform two types of operations:

Subtract 1 from x. This operation costs you A coins.
Divide x by k. Can be performed only if x is divisible by k. This operation costs you B coins.

What is the minimum amount of coins you have to pay to make x equal to 1?

Input

The first line contains a single integer n (1 ≤ n ≤ 2·109).

The second line contains a single integer k (1 ≤ k ≤ 2·109).

The third line contains a single integer A (1 ≤ A ≤ 2·109).

The fourth line contains a single integer B (1 ≤ B ≤ 2·109).

Output

Output a single integer — the minimum amount of coins you have to pay to make x equal to 1.

Examples
inputCopy
9
2
3
1
output
6
inputCopy
5
5
2
20
output
8
inputCopy
19
3
4
2
output
12

题意：给你一个n，让你把n变成1，每次有两种操作：减一花费A，除以k花费B（整除的前提下），问你最小花费。

思路：简单贪心，当不能整除时变成整除，此时只能减。当整除时，判断除的花费和减的花费，注意特判k==1的情况，防止死循环，因为这个TLE了一发。。。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,k,a,b,ans;
int main()
{
    while(~scanf("%lld%lld%lld%lld",&n,&k,&a,&b))
    {
        if(k>n||k==1){printf("%lld\n",a*(n-1));continue;}
        ans=0;
        while(1)
        {
            if(n%k!=0)     //不能整除只能减法，减到整除为止
            {
                ans+=a*(n%k);
                n=n-(n%k);
            }
            if(n<k)   //此时只能减法，除法永远用不到，直接求解答案
            {
                ans+=a*(n-1);
                break;
            }
            if(n==1)break;
            ans+=min(a*(n-n/k),b);  //减法和除法取最小花费
            n/=k;
            if(n==1)break;
        }
        printf("%lld\n",ans);
    }
    return 0;
}