40 lines
2.3 KiB
Rexx
40 lines
2.3 KiB
Rexx
/*REXX program solves the knapsack/unbounded problem: highest value, weight, and volume.*/
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maxPanacea=0 /* value weight volume */
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maxIchor =0 /* ═══════ ═══════ ══════ */
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maxGold =0; panacea.$ = 3000 ; panacea.w = 0.3 ; panacea.v = 0.025
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max$ =0; ichor.$ = 1800 ; ichor.w = 0.2 ; ichor.v = 0.015
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now. =0; gold.$ = 2500 ; gold.w = 2 ; gold.v = 0.002
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# =0; sack.$ = 0 ; sack.w = 25 ; sack.v = 0.25
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L =0
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maxPanacea= min(sack.w / panacea.w, sack.v / panacea.v)
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maxIchor = min(sack.w / ichor.w, sack.v / ichor.v)
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maxGold = min(sack.w / gold.w, sack.v / gold.v)
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do p=0 to maxPanacea
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do i=0 to maxIchor
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do g=0 to maxGold
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now.$ = g * gold.$ + i * ichor.$ + p * panacea.$
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now.w = g * gold.w + i * ichor.w + p * panacea.w
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now.v = g * gold.v + i * ichor.v + p * panacea.v
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if now.w > sack.w | now.v > sack.v then iterate i
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if now.$ > max$ then do; #=0; max$=now.$; end
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if now.$ = max$ then do; #=#+1; maxP.#=p; maxI.#=i; maxG.#=g
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max$.#=now.$; maxW.#=now.w; maxV.#=now.v
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L=max(L, length(p + i + g) )
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end
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end /*g (gold) */
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end /*i (ichor) */
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end /*p (panacea)*/
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L=L + 1
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do j=1 for #; say; say copies('▒', 70) "solution" j
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say ' panacea in sack:' right(maxP.j, L)
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say ' ichors in sack:' right(maxI.j, L)
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say ' gold items in sack:' right(maxG.j, L)
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say '════════════════════' copies("═", L)
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say 'carrying a total of:' right(maxP.j + maxI.j + maxG.j, L)
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say left('', 40) "total value: " max$.j / 1
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say left('', 40) "total weight: " maxW.j / 1
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say left('', 40) "total volume: " maxV.j / 1
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end /*j*/
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/*stick a fork in it, we're all done. */
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