A game programming problem urgent!! ）

Indeed, this ** amount is not something that can be done by dozens of lines, it belongs to commercial applications, so let's spend money to buy it.

Dude, this is too much! I feel like I'm back in time when I wrote Tetris!

Is there any money?

This one is still more difficult.

Drawer principle problem.

Divide the 1-meter-long line into 5 equal segments.

The length of each segment is 20 cm.

Put 6 points into these 5 segments, and at least 2 points should be placed in the same segment, so that the distance between the two points should not exceed 20 cm.

Counter-argument, if any two points are 20 centimeters larger.

There are 5 intervals between 6 points, all of which are greater than 20, so the total length is greater than 100.

There are 5 intervals in the middle of these 6 points, assuming that each interval is greater than 20 cm, then the total distance is greater than 20 5 = 100 cm = 1 meter, so the assumption is not true, so at least one interval is not more than 20 cm.

Profit and loss issues.

The difference in the total number of rounds is 680-200 = 480 rounds, the difference in the number of rounds carried by each person is 50-45 = 5 rounds, the number of rounds is 480, 5 = 96 people, and the bullets are 50 96 200 = 5000 rounds.

The double profit problem in the profit and loss problem.

45 rounds per person, 680 rounds more (surplus).

50 rounds per person, 200 rounds more (surplus).

Number of people: (680-200) (50-45) = 96 peopleBullets: 96 45 + 680 = 5000 rounds.

If each person carries 45 rounds, it will be 680 more rounds; If each person carries 50 rounds, there are 200 more rounds, so if each person carries 50-45 rounds more, and the remaining 680-200 = 480 rounds, then there are 480 enemies 5 = 96 people.

There are bullets 50 * 96 + 200 = 5000 rounds.

Let x people y bullets, there is a system of equations:

45x+680=y

50x+200=y

Solution: x = 96 people y = 5000 rounds.

If everyone carries the same number of bullets, there are (680-200) more bullets than 45 rounds on their backs, and if everyone carries the same number of bullets, there will be (680-200) (50-45) = 96 enemies, and the bullets are 96*45+680=5000 bullets.

This is a classic extreme problem, and the basic idea is this:

It is possible to touch 4 pairs of the same color at one time, but this can not happen at any time, so consider the worst, that is, 3 pairs of red, yellow, black, and white are touched, and at this time, as long as you touch any pair of socks of the same color, there are 4 pairs of socks of the same color!

So find at least 13 pairs of socks to ensure that the purpose is achieved!

There is less to consider the most extreme cases.

Each type has 3 pairs, which need to be touched:

4 3 = 12 pairs.

Now as long as you find another pair, you can ensure that there are at least 4 pairs of the same color: 12 + 1 = 13 pairs.

Let's find three pairs each. Touched it 12 times, and touched it again for sure.

1*2 is to add from 1 to the second number.

2*3 is to add from 2 to the 3rd number.

And so on.

1。A philatelic enthusiast bought 100 stamps of 10 cents and 20 cents, with a total value of 18 yuan and 8 jiao. Does this philatelic enthusiast have a few stamps of each of these two types?

Solution: The stamps of 10 points are:

= 12 sheets. 20 stamps are: 100-12=88 sheets.

2.The school bought 3 volleyballs and 2 soccer balls, which cost a total of 111 yuan. Each soccer ball is 3 yuan more expensive than each volleyball. How much is each for each volleyball and each soccer ball?

Solution: Every soccer ball is:

111+3*3) (3+2)=24 yuan.

Each volleyball is: 24-3=21 yuan.

3.The price of 2 fountain pens is the same as the price of 8 ballpoint pens. If you buy 3 fountain pens and 5 ballpoint pens for a total of 17 yuan, how much is each of the two types of pens?

Solution: The price of 2 fountain pens is equal to the price of buying 8 ballpoint pens, so the price of 1 fountain pen is equal to the price of buying 4 ballpoint pens;

Ballpoint pen**: 17 (3*4+5)=1 yuan.

Fountain pen**: 1*4=4 yuan.

1. Solution: Suppose 100 stamps are 10 points, 10 100 = 1000 points.

1000 1880, indicating that there are 20 cents of stamps out of 100 stamps, and 20 cents of stamps are exchanged for 10 cents of stamps).

Then the number of stamps of 20 points (1880 1000) (20 10) = 88 stamps of 10 points The number of stamps 100 88 = 12 stamps 2, Solution: Assuming that the 5 balls bought by the school are footballs, then there is.

** (111 3 3) 5 = 24 yuan for volleyball** 24 3 = 21 yuan.

3. Solution: Suppose the pens you buy are ballpoint pens, because the price of one fountain pen is equal to the price of four ballpoint pens, so.

The price of a ballpoint pen 17 (3 4 5) = 1 yuan The price of a fountain pen 1 4 = 4 yuan.

As can be seen from the question, 743 is a group of three numbers in a loop.

So 100 3 = 33 groups with 1 remaining number.

So the 100th digit is 7