2. The first turn is necessary, but the second turn will reduce the number. You have to walk 400 kilometers for the first time, and then go back and install it. The second time is also a repetition. You can put down 200 bananas at a time, which is 400 bananas. . Then there is no need to look back for the third time. 1000 bananas were loaded, and 600 bananas were left after 400 kilometers. Just put the original 400 bananas on, just 1000 bananas covered the remaining 600 kilometers, and finally there were 400 bananas left, which should be the most.
3. Minimize the round trip. 300 meters, 400 drops, three round trips. At 300m, there is1200 at the moment; 350 meters, down 500, twice round trip. There is 1000 at 350m, leaving 350.
4. Exclusion method, setting the turning point at 1000m is not feasible. 1000m sets two turning points, A and B, that is, 1000m is divided into three sections. Let the length of each segment be X 1 from the starting point to point A, X2 from point A to point B, and X3 from point B to the end point.
X 1+X2+X3= 1000
It takes five times to transport 3000 bananas to point A. Then there are 3000-5 * x/kloc-0 bananas at point A,
Through reasoning (details omitted), 3000-5*X 1 is less than or equal to 2000,
In this way, to point B, 3000-5*X 1-3*X2, this value is less than or equal to 1000.
The finish line is 3000-5*X 1-3*X2-X3. According to the above judgment, if all people are equal:
x 1 = 200； X2=334
Finally, there are 532 left.
5. First of all, every 2.5 round trips from 0 to 0~200km, all bananas are moved to the lower part 1 km, * * * consumes 1000 bananas, leaving 2000 bananas.
Secondly, every 1km 1.5 times from 200 to 533 km, all bananas are transported to the next 1km in turn, and 999 bananas are consumed, leaving 100 1.
Finally: 533~ 1000km, carrying 1000 bananas to the destination (leaving one at 533), eating 467 bananas on the way, leaving 1000-467=533 bananas.
6, divided into three stages, 1, three times 2000 to 200 kilometers; Carrying 1000 vehicles to advance 333 kilometers twice; Finally, take/kloc-0.000 bananas, walk the remaining 467 kilometers, and save 533 bananas.
7. The first time: take 1000 bananas for 250 kilometers, leave 500 bananas, come back and eat the remaining 250 bananas and walk 500 kilometers.
The second time: take 1000 bananas and walk 500 kilometers, leaving 250 bananas, and return to 250 kilometers. Take 250 bananas from the remaining 500 bananas here and return, and walk 1000 kilometers.
The third time, take the last 1000 bananas for 250 kilometers, take the 250 bananas here, continue to walk for 500 kilometers, take the 250 bananas here, continue to walk for the remaining 500 kilometers, and finally leave 500 bananas for 1000 kilometers.
A total of 2500 kilometers, leaving 500 bananas.
8. Take 1000 at most at a time, so you need to run between 200 1 and 3000 for 5 times (plus the number of round trips twice). If the number of pieces is less than 2000, it is not cost-effective to run five times. You need to run for 3 times in 2000 and below, 1000 and below 1 time, and you can run for 200 times in one breath for the first time (3000-5 * 200 = 2000). If you can't run 200, it doesn't matter. Run a few more times and the effect will be the same. Run 333 for the second time (2000-3 * 333 =1000) (unnecessary). You can take 1000 for the third time and only run once. To the finish line, it can be1000-(1000-200-333) = 533.
9. I just thought about going back and forth five times, not taking 1000 yuan at a time. In other words, if it's only 2000, just walk back and forth three times.
(1) Take 1000, walk 200 kilometers, put down 600, and return to point A with only 0; Take 1000, walk 200 kilometers, put down 600, and return to point A with only 0; Then take the last 1000 to 200 kilometers and put down 800. There are 2000 * * at 200 kilometers.
(2) Take 1000 block, walk from 200 km to 533 km, put down 334 blocks, and return to 200 km with 0 blocks left; Take the last 1000, walk from 200 km to 533 km, and put down 667. Now there is 100 1 at 533km. Take 1000, walk from 533km to point B, and take 533.
10, bring as many bananas as possible in front, and go straight ahead when there is only 1000 in the back, which will consume the least. When there are many belts ahead, it is not important to set a turning point. As long as it is not too big, it is basically the same to go back every kilometer and two kilometers. Take more bananas according to the progress per kilometer. At first, all bananas need to be transported twice per kilometer, and after 200 kilometers, there are 2,000 bananas left. You only need to turn back three bananas per kilometer, and then 333 kilometers, leaving only 100 1 (more 1, give it to the camel when you leave).
1 1, I think so. No matter how many stops AB needs to get the best answer, the key point is:
* * * * * * * No matter which station the camel starts from, it is most efficient to run to the next station (or as close as possible) with full load. And the higher the efficiency, the more bananas can be saved.
Let x = A reach the stop point 1.
1, there are 3000 bananas, which means that the camel must run three times at the first stop (the third stop does not need to return), that is, the value of the first stop 1 remaining bananas must be: (1000-2x) * 2 (1000-x) = 30.
2. 1, assuming there are only 1 stop points, as I said just now, the efficiency is the highest when x=400 (because there are 1 000 bananas left in the stop point), that is, in the case of only one stop point, there are still 400 bananas left in the end.
Suppose there are two stop points, and let y = stop point 1 to stop point 2.
2.2, x=200 is the most efficient. At this time, there are 2000 bananas (that is, 2000 bananas) left at the stop point of 1. At this time, the problem becomes that 2000 bananas run 800 meters. Then the number of bananas left after stop 2 becomes: (1000-2y) (1000-y) = 2000-3y.
2.2. 1, according to the conclusion from point A to stop point 1, the efficiency is the highest when y=333, because there are 100 1 bananas left in stop point 2 at this time. (That is, I ran 467 meters with 1000 bananas and finally saved 533 bananas.)