Given 3 integers K, P and N. Where, K is the number of problems given to the person every day and P is the maximum number of problems he can solve in a day. Find the total number of problems not solved after the N-th day.
Input : K = 2, P = 1, N = 3 Output : 3 On each day 1 problem is left so 3*1 = 3 problems left after Nth day. Input : K = 4, P = 1, N = 10 Output : 30
If P is greater than or equal to K then all problems will be solved on that day or (K-P) problems will be solved on each day so the answer will be 0 if K<=P else the answer will be (K-P)*N.
Below is the implementation of the above approach:
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