### Project Euler 201 Subsets with a unique sum

For any set A of numbers, let sum(A) be the sum of the elements of A.

Consider the set B = {1,3,6,8,10,11}.

There are 20 subsets of B containing three elements, and their sums are:

Consider the set B = {1,3,6,8,10,11}.

There are 20 subsets of B containing three elements, and their sums are:

sum({1,3,6}) = 10,

sum({1,3,8}) = 12,

sum({1,3,10}) = 14,

sum({1,3,11}) = 15,

sum({1,6,8}) = 15,

sum({1,6,10}) = 17,

sum({1,6,11}) = 18,

sum({1,8,10}) = 19,

sum({1,8,11}) = 20,

sum({1,10,11}) = 22,

sum({3,6,8}) = 17,

sum({3,6,10}) = 19,

sum({3,6,11}) = 20,

sum({3,8,10}) = 21,

sum({3,8,11}) = 22,

sum({3,10,11}) = 24,

sum({6,8,10}) = 24,

sum({6,8,11}) = 25,

sum({6,10,11}) = 27,

sum({8,10,11}) = 29.

sum({1,3,8}) = 12,

sum({1,3,10}) = 14,

sum({1,3,11}) = 15,

sum({1,6,8}) = 15,

sum({1,6,10}) = 17,

sum({1,6,11}) = 18,

sum({1,8,10}) = 19,

sum({1,8,11}) = 20,

sum({1,10,11}) = 22,

sum({3,6,8}) = 17,

sum({3,6,10}) = 19,

sum({3,6,11}) = 20,

sum({3,8,10}) = 21,

sum({3,8,11}) = 22,

sum({3,10,11}) = 24,

sum({6,8,10}) = 24,

sum({6,8,11}) = 25,

sum({6,10,11}) = 27,

sum({8,10,11}) = 29.

Some of these sums occur more than once, others are unique.

For a set A, let U(A,k) be the set of unique sums of k-element subsets of A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and sum(U(B,3)) = 156.

For a set A, let U(A,k) be the set of unique sums of k-element subsets of A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and sum(U(B,3)) = 156.

Now consider the 100-element set S = {1

S has 100891344545564193334812497256 50-element subsets.

^{2}, 2^{2}, ... , 100^{2}}.S has 100891344545564193334812497256 50-element subsets.

Determine the sum of all integers which are the sum of exactly one of the 50-element subsets of S, i.e. find sum(U(S,50)).

I solved this problem for 2hours coding.

I was'nt able to solve this using brutefoce, so I made algorithm.

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