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