There should be max 334 bananas to be carried to destination.
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Kejeg
2009-07-15 11:15
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骆驼香蕉问题解决办法:
1. From A to find out point C
3000 - (2X + 2X +X) = 2000
So X = 1000/5 = 200
2. From C to find D
2000 - (2Y + Y) = 1000
So Y = 1000/3 = 333.33
3. From D to B = 1000 - 200 - 333 = 467
So 骆驼 can carry to B 1000 - 467 = 533 香蕉.
1 香蕉 is left at point D.
Detail description:
The route is:
A(0) -> C(200) -> D(533) -> B(1000)
1. From A to C, 骆驼 carry 3 times. First time leave 600 at C. Second time leave 600 at C. Third time leave 800 at C. That's 2000.
2. From C to D, 骆驼 carry 2 times. First time leave 334 at D. Second time leave 667 at D. That's 1001.
3. From D to B, 骆驼 leave 533 to B. And 1 香蕉 is left at point D.
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