Prerequisites : DP , Maths
Difficulty : Medium
Expected Complexity : O(N*M + T).
Editorial :
The very first thing to observe is that since each of the apple received is distinct , one of the ways is surely going to be of the form : a1 < a2 < a3 < ... < an , where ai denotes apples received by ith shinigami. If we find the number of ways to arrange this , we can multiply this by N! to find all possible distinct ways. So our problem reduces to finding the number of ways to distribute M apples such that a1 + a2 + a3 + ... + an = M and 0 < a1 < a2 < ... < an.
Let's rewrite each ai as ai = a(i-1) + 1 + yi , where yi >= 0.
a1 = 1 + y1;
a2 = a1 + 1 + y2;
a3 = a2 + 1 + y3;
...
an = a(n-1) + 1 + yn;
Replacing all a(i-1) in ai , we get :
a1 = 1 + y1;
a2 = 2 + y1 + y2;
a3 = 3 + y1 + y2 + y3;
....
an = n + y1 + y2 + ... + yn;
summation of all ai = 1 + 2 + 3 + .. + n + n*y1 + (n-1) * y2 + (n-2) * y3 + ... yn;
renaming our yi as y(n-i+1) , we can get :
sum = n*(n+1)/2 + summation i*yi;
M - n*(n+1)/2 = summation i*yi;
now if lhs < 0 , then answer is 0.
otherwise K = summation i*yi , where yi >= 0;
we can easily solve this using the recurrence : solve(i,K) = solve(i-1,K) + solve(i,K-i).
In short, the answer is : N! * solve(n , m - n*(n+1)/2).