34 lines
1.5 KiB
Text
34 lines
1.5 KiB
Text
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T Bounty
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Int value
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Float weight, volume
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F (value, weight, volume)
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(.value, .weight, .volume) = (value, weight, volume)
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V panacea = Bounty(3000, 0.3, 0.025)
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V ichor = Bounty(1800, 0.2, 0.015)
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V gold = Bounty(2500, 2.0, 0.002)
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V sack = Bounty( 0, 25.0, 0.25)
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V best = Bounty( 0, 0, 0)
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V current = Bounty( 0, 0, 0)
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V best_amounts = (0, 0, 0)
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V max_panacea = Int(min(sack.weight I/ panacea.weight, sack.volume I/ panacea.volume))
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V max_ichor = Int(min(sack.weight I/ ichor.weight, sack.volume I/ ichor.volume))
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V max_gold = Int(min(sack.weight I/ gold.weight, sack.volume I/ gold.volume))
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L(npanacea) 0 .< max_panacea
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L(nichor) 0 .< max_ichor
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L(ngold) 0 .< max_gold
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current.value = npanacea * panacea.value + nichor * ichor.value + ngold * gold.value
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current.weight = npanacea * panacea.weight + nichor * ichor.weight + ngold * gold.weight
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current.volume = npanacea * panacea.volume + nichor * ichor.volume + ngold * gold.volume
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I current.value > best.value & current.weight <= sack.weight & current.volume <= sack.volume
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best = current
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best_amounts = (npanacea, nichor, ngold)
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print(‘Maximum value achievable is ’best.value)
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print(‘This is achieved by carrying (one solution) #. panacea, #. ichor and #. gold’.format(best_amounts[0], best_amounts[1], best_amounts[2]))
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print(‘The weight to carry is #2.1 and the volume used is #.3’.format(best.weight, best.volume))
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