标签：from ring cin and ati put algo orm min
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:
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 a single integer — the minimum amount of coins you have to pay to make x equal to 1.
9
2
3
1
6
5
5
2
20
8
19
3
4
2
12
In the first testcase, the optimal strategy is as follows:
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.
#include <iostream> #include <vector> #include <algorithm> #include <string> #include <set> #include <queue> #include <map> #include <sstream> #include <cstdio> #include <cstring> #include <numeric> #include <cmath> #include <unordered_set> #include <unordered_map> //#include <xfunctional> #define ll long long #define mod 1000000007 using namespace std; int dir[4][2] = { { 0,1 },{ 0,-1 },{ -1,0 },{ 1,0 } }; const long long INF = 0x7f7f7f7f7f7f7f7f; const int inf = 0x3f3f3f3f; int main() { ll n, k, A, B; ll res=0; cin >> n >> k >> A >> B; if (k == 1) cout << (n - 1)*A; else { while (n != 1) { if (n < k) { res += (n - 1)*A; n = 1; } else { if (n%k > 0) { res += (n%k)*A; n -= n%k; } if ((n - n / k)*A > B) { res += B; n /= k; } else { res += (n - n / k)*A; n /= k; } } } cout << res; } return 0; }
标签：from ring cin and ati put algo orm min
原文地址：https://www.cnblogs.com/dealer/p/12442701.html
