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
9
2
3
1
Output
6
Input
5
5
2
20
Output
8
Input
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.
题目链接
https://vjudge.net/contest/280107#problem/B
题目大意
通过两种方式将数n变为1
- 除以k，花费b枚硬币
- 减1，花费a枚硬币
数据范围
n (1 ≤ n ≤ 2·1e9)
k (1 ≤ k ≤ 2·1e9)
A (1 ≤ A ≤ 2·1e9)
B (1 ≤ B ≤ 2·1e9)
解题思路
首先要将n 用第二种方式变为 k 的倍数，每次计算直接n / k 一步到位，用n-- 会超时，特判一下k == 1。以后做题需要注意观察数据范围，避免超时。
注意
如果用ans记录答案，不能先对n进行变换，要记录ans后，再对n进行操作
解决代码
#include<cstdio>
int main()
{
long long n,k,a,b,ans = 0;
scanf("%lld %lld %lld %lld",&n,&k,&a,&b);
while(n != 1){
if(k == 1){
ans = (n - 1) * a;
break;
}
if(n % k != 0){
ans = ans + (n % k) * a;
n = n - n % k;
}
if((n - n / k ) * a >= b&&n % k == 0)
{
n = n / k;
ans += b;
}
else {
ans = ans + (n - n / k) * a;
n = n / k;
}
if(n < k) {
ans = ans + (n - 1) * a;
break;
}
}
printf("%lld\n",ans);
return 0;
}