Short Problem Definition:
Little Bob loves chocolates, and goes to a store with $N in his pocket. The price of each chocolate is $C. The store offers a discount: for every M wrappers he gives to the store, he gets one chocolate for free. How many chocolates does Bob get to eat?
Link
Complexity:
time complexity is O(N);
space complexity is O(1)
Execution:
Evaluate the number of wraps after each step. Do this until you have enough wraps to buy new chocolates.
Solution:
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#include <cmath>#include <cstdio>#include <vector>#include <iostream>#include <algorithm>using namespace std;int eatenChocolades(int availableCash, int price, int wrapperDiscount){ int eaten = 0; int wraps = 0; wraps = eaten = availableCash/price; while (wraps >= wrapperDiscount){ int newlyEaten = wraps/wrapperDiscount; eaten += newlyEaten; wraps %= wrapperDiscount; wraps += newlyEaten; } return eaten;}int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int t,n,c,m; cin>>t; while(t--){ cin>>n>>c>>m; int answer=0; answer = eatenChocolades(n,c,m); cout<<answer<<endl; } return 0;} |
