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Thursday, 17 september 2015

troubles with recipients

Proposed by catanedelcuOffline
(4 comments)1.667 times displayed

Trei recipiente R0, R1 si R2 contin respectiv volumele de lichid  v0 ≥ v1 ≥ v2 ≥1. 

Volumul total al fiecarui recipient este V > v0+v1+v2.  Cu alte cuvinte, oricare recipient poate cuprinde toate cele trei volume de lichid la un loc.

Regula:  dintr-un recipient se poate turna lichid in alt recipient numai daca volumul aflat in cel din urma se dubleaza.

Problema:  sa se arate ca pentru orice volume  v0, v1, v2 (numere naturale mai mari ca 1) , exista o cale de a goli unul dintre recipiente, respectand regula, intr-un numar finit de pasi.

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Pentru incalzire se propune rezolvarea unui caz particular pentru volumele initiale 17, 8, 5.


Tags: matrh
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