SarK0Y

Amr Morsi, you said about algo for getting of MIN(0)? Merged post follows: Consecutive posts mergedhmm.. i did let wrong moment:-) int x=M(n-1), y=M(n-2);

Amr Morsi, thanks for sharing your idea. i got MIN(0) with following algo: //(given: set M with n integers) sort(M); //M(0)<M(1)..<M(n-1) int x=n-1, y=n-2; for(int i=n-3; i>-1; i--) { if(x>y)y*=M(i); else x*=M(i); }

Amr Morsi, prime[general] task is to get common algo.

You're right. Merged post follows: Consecutive posts mergedi got idea how to get MIN(0), but other iteration is darkness for me:-(

yeah, i'm follower of balance's ideology too:-) however, this ideology leads us to religious conception. Mr Skeptic, that's no about kinetic/potential energy, that's about what is energy per se & where is source of energy.

problem can be rephrased so: [math]\prod_{f\in M}f>x[/math], M is given set, x is given number & [math]\prod_{f\in M}f-x [/math] must have minimal value among possible solutions.

Good Time to All, Amici. i got following math problem: to split up number array into two subsets (S1, S2) where X[t] & Y[t] are multiplications of numbers of S1 & S2 respectively, t is index of iteration with condition: |X(0)-Y(0)|==MIN(0), |MIN(0)-(X(1)-Y(1))|==MIN(1), ..., |MIN(n-1)-(X(n)-Y(n))|==MIN(n). for example, let's take an array: 2, 5, 7 then: first pair is {2, 5}, {7} ==> X(0)==2*5==10, Y(0)==7, MIN(0)==3; second is {2, 7}, {5} ==> X(0)==14, Y(0)==5, MIN(1)==6; third is .. ------------------------------------------------ Thanks a lot in Advance.