CF940B Our Tanya is Crying Out Loud

最新推荐文章于 2021-08-05 22:38:45 发布

转载最新推荐文章于 2021-08-05 22:38:45 发布 · 117 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/l609929321/p/8469391.html

本文介绍了一个数学问题：通过两种操作（减1或除以k）将整数n变为1的最小成本。文章提供了详细的解析思路及C++实现代码。

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

input

Copy

output

input

Copy

output

input

Copy

output

Note

In the first testcase, the optimal strategy is as follows:

Subtract 1 from x (9 → 8) paying 3 coins.
Divide x by 2 (8 → 4) paying 1 coin.
Divide x by 2 (4 → 2) paying 1 coin.
Divide x by 2 (2 → 1) paying 1 coin.

The total cost is 6 coins.

In the second test case the optimal strategy is to subtract 1 from x 4 times paying 8 coins in total.

题目的意思是给你两种操作，一种是每次减1消耗a元，一种是每次除以k每次消耗b元。使n变成1的最小消耗。

在每次没有达到k的倍数前，只能减一，达到后判断下消耗是减去还是除以小，选小的。

#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define maxn 100010
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
int main() {
    ll n,k,a,b;
    while( cin >> n >> k >> a >> b ) {
        ll ans = 0;
        if( k == 1 ) {
            cout << ( n - 1 ) * a << endl;
            continue;
        }
        while( n != 1 ) {
             if( n % k == 0 ) {
                ans += min( ( n - n / k ) * a, b );
                n /= k;
             } else if( n > k ) {
                ans += ( n % k ) * a;
                n -= n % k;
             } else {
                ans += ( n - 1 ) * a;
                n = 1;
             }
        }
        cout << ans << endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/l609929321/p/8469391.html