Codeforces Round #466 (Div. 2) B. Our Tanya is Crying Out Loud[將n變爲1，有兩種方式,求最小花費/貪心]

B. Our Tanya is Crying Out Loud

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. Subtract 1 from x. This operation costs you A coins.
2. 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

9
2
3
1

Output

6

Input

Copy

5
5
2
20

Output

8

Input

Copy

19
3
4
2

Output

12

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.

[題意]:將n變爲1，有兩種方式① n-1花費a , ② 若是n可以被k整除，n=n/k,花費b.求最小花費.

[分析]:很顯然這是道貪心題，
一個很容易錯的地方就是想着優先除二，而後在減一，
由於沒有考慮二者的花費，若是b很大的話就錯了，因此每次都要判斷下，而後注意細節就好了

首先若是k==1，那麼ans=(n-1)*a;

n%k!=0,只能執行操做1

n<k,只能執行操做1
n%k==0(n>=k) 的狀況下比較n到達相同結果時兩種操做的花費便可。

[代碼]:

/* 題意：將n變爲1，有兩種方式①n-1花費a,②若是n可以被k整除，n=n/k,花費b,求最小花費 */
/* 很顯然這是道貪心題， 一個很容易錯的地方就是想着優先除二，而後在減一， 由於沒有考慮二者的花費，若是b很大的話就錯了，因此每次都要判斷下，而後注意細節就好了 */ #include<bits/stdc++.h>
using namespace std; #define pb push_back
#define mem(a) memset(a,0,sizeof(a)) typedef long long ll; typedef pair<int,int> pii; const int maxn=150; const int inf=0x3f3f3f3f; int main() { ll x,n,k,a,b,ans; while(~scanf("%lld%lld%lld%lld",&n,&k,&a,&b)){ ans=0; if(k==1) return 0*printf("%lld\n",a*(n-1));  //注意特判k==1的狀況，防止死循環
        while(n!=1){ //n沒法整除k
            if(n%k!=0){    //不能整除只能減法(a)，減到整除爲止 
                ans+=a*(n%k); n-=(n%k); } if(n<k){       //此時只能減法(a)，除法永遠用不到，直接求解答案 
                ans+=a*(n-1); break; } //n能夠整除k
            ans+=min(b,a*(n-n/k)); //減法(a)和除法(b)取最小花費 
            n/=k; //n能夠整除k
 } printf("%lld\n",ans); } } /* n-n/k*k = n%k 獲取距離n最近的k的倍數的方法是:n/k*k = n-n%k */

貪心