**Making Candies**

### Problem Statement :

Karl loves playing games on social networking sites. His current favorite is CandyMaker, where the goal is to make candies. Karl just started a level in which he must accumulate n candies starting with m machines and w workers. In a single pass, he can make m * w candies. After each pass, he can decide whether to spend some of his candies to buy more machines or hire more workers. Buying a machine or hiring a worker costs p units, and there is no limit to the number of machines he can own or workers he can employ. Karl wants to minimize the number of passes to obtain the required number of candies at the end of a day. Determine that number of passes. Function Description Complete the minimumPasses function in the editor below. The function must return a long integer representing the minimum number of passes required. minimumPasses has the following parameter(s): m: long integer, the starting number of machines w: long integer, the starting number of workers p: long integer, the cost of a new hire or a new machine n: long integer, the number of candies to produce Input Format A single line consisting of four space-separated integers describing the values of m, w , p , and n, the starting number of machines and workers, the cost of a new machine or a new hire, and the the number of candies Karl must accumulate to complete the level. Constraints 1 <= m, w, p, n <= 10^12 Output Format Return a long integer denoting the minimum number of passes required to accumulate at least n candies. Sample Input 3 1 2 12 Sample Output 3

### Solution :

` ````
Solution in C++ :
In C ++ :
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool check(ll machines, ll workers, ll price, ll target, ll rounds) {
if (machines >= (target+workers-1)/workers) return true;
ll cur = machines*workers;
rounds--;
if (rounds == 0) return false;
while (1) {
ll rem = target - cur;
ll rnds = (rem + machines*workers - 1) / (machines*workers);
if (rnds <= rounds) return true;
if (cur < price) {
rem = price - cur;
rnds = (rem + machines*workers - 1) / (machines*workers);
rounds -= rnds;
if (rounds < 1) return false;
cur += rnds * machines * workers;
}
cur -= price;
if (machines > workers) {
workers++;
} else {
machines++;
}
}
return false;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
ll m, w, p, n;
cin >> m >> w >> p >> n;
ll a = 1, b = 1000000000000LL;
while (a < b) {
ll mid = (a + b) >> 1;
if (check(m, w, p, n, mid)) {
b = mid;
} else {
a = mid + 1;
}
}
cout << a << "\n";
return 0;
}
```

` ````
Solution in Python :
In Python3 :
def minimumPasses(m, w, p, n):
candy = 0
invest = 0
spend = sys.maxsize
while candy < n:
passes = (p - candy) // (m * w)
if passes <= 0:
mw = (candy // p) + m + w
half = math.ceil(mw / 2)
if m > w:
m = max(m, half)
w = mw - m
else:
w = max(w, half)
m = mw - w
candy %= p
passes = 1
candy += passes * m * w
invest += passes
spend = min(spend, invest + math.ceil((n - candy) / (m * w)))
return min(invest, spend)
```

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