Java written question: Camel water and car oil problem. (My own opinion, hope everyone good, correct me)

A businessman with Donkey Yun Carrot 3000 to sell, desert long 1000 kilometers, donkey a maximum camel 1000, but each walk a kilometer to eat a carrot, ask the businessman can sell how many carrots.

Solution: Set to transport X meters to return to the transport, then transported to X 3 times, X at the Radish 3 (1000-2x);

where x must be less than 500.

The rest of the way: 1000-x, assuming that the middle does not stop a trip: Then 3 (1000-2x) >=1000, x<=1000/3; to reach the destination of up to 1000/3 left;

Suppose two times: 1000-x is greater than 500 and cannot be back 2 times. Loft

If the remaining points 2 times, then from the know must be at the set Y stop, you can see X to y at least 2 times, y to the destination 1 trip, 3 (1000-2x) >=2000 then X=1000/6, 2 (1000-2y) =1000,y=1000/ 2 So the arrival destination remains: 1000-X-Y=1000-1000/6-1000/2=1000/3;

If divided 4 times, the last time must be a trip, then the penultimate second time is two times, analogy for the first time for 4 times, impossible, so up to 3 times to go.

So the answer is 1000/3.

Change the subject

The journey is 1000, the origin of the carrot is countless, asked to arrive at the destination of 1000, at least how much carrot consumption?

Suppose 3 times (average distance):

Last: (1000-2/3*1000)/(1000/3) + 1 trips = 2

Countdown two times: (2000-2/3*1000)/(1000/3) +1=5

Countdown three times: 13

So 13000 root.

Suppose N times

Last an= (1/n)/(1-2/n) +1=1/(n-2) +1 so n can only be 3

The following is a discussion of the non-average distance (time relationship, next!) ）

Java written question: Camel water and car oil problem. (My own opinion, hope everyone good, correct me)