1 1 platform ladder

How much is APM? How many soldiers do you make up every minute? How is the economic situation? It all counts. Add 87 points, that must be an extraordinary performance. . . . Don't add so much when the steps are similar to your level.

The ladder is compared with yourself, not with others.

1 1 step integral system: the hero weight coefficient is similar to the normal probability distribution. The correction coefficient is a convergence function with a limit of 1. Step integral =∑PmXmSm, ∑P= 1, p is the weight, s is the correction coefficient (0.75~ 1), and x is the hero integral of a single hero.

Under the above principles, everything is easier to explain.

Temporarily set your ladder at 1500.

1500 gives all your heroes a weighted average. That is ∑PmXm= 1500. ∑P= 1。

Suppose you hit 29 heroes, then m is 1~29, and p is the weight, ignoring the correction coefficient.

Next, suppose you hit another hero: the fire girl. And hit 1700, do you think 1700- 1500=200?

Obviously wrong. Suppose the fire girl's Pn=0.02, then the original ∑Pm changes from 1 to 0.98, and your average hero score actually becomes1700 * 0.02+1500 * 0.98 (here, for the convenience of calculation, I regard each hero score of your top 29 heroes as/. This is the last step integral you can add.

Then you will ask? Then I only played five heroes before. Why did I add 100 to each set? There is a simple reason.

Suppose you play four heroes, ∑PnXn= 1200, ∑P= 1. When you play the fifth hero, you get 1700 points for your outstanding performance, assuming that the weight of this hero is 0.2, 1700 * 0.2+ 17. That's why you can get 100 points at this time.

What if you hit 1 hero? According to the above theory, if the hero score is 1700, isn't that the ladder score of 1700? In fact, when you only hit one hero, the first set only gives you a correction coefficient 1700* (this coefficient is actually in the above formula, which is determined according to the number of times you hit a hero and the wide number, which is about 0.75 approaching 1. The wider you play, the greater the correction coefficient. I think this coefficient is similar to convergence to 1. That is to say, if your first hero killed 40 people and 5 people in the first set, your hero score is 2000 points directly, then multiplied by the correction factor, your ladder score will reach 1500 points in the first set.

Next, you won't get more points if you play with high-scoring heroes. For example, after you play a set of +3 hero points, the hero points will reach 2003 points. (2003*0.5+2000*0.5-2000)*0.76 is about = 1, but if you lose a set, suppose you lose a set of hero points, and deduct 200 to become 1800 points, then: (1800 *.

That's why I played 1 high-scoring hero, and once I lost, I would buckle wildly.

If you hit a lot of heroes, suppose it's 30. One of the high-scoring heroes lost, assuming that the hero's score was 1700, the hero's score after deducting 300 points became 1400, and your original ladder score was exactly 1700. So how many ladder points do you want to deduct? Suppose your dragon's weight coefficient p is 0.03, then you should deduct -300*0.03=-9 points.

Then you asked again. . If I hit 90 heroes, how can I calculate the score of the 90th hero?

It's simple. In the ladder integral system, the hero weight coefficient P after the top 30 heroes is getting smaller and smaller (just like normal distribution). If your 90-level hero originally scored 800 points and lost a set and deducted 200 points, but at this time the hero's weight is 0.002 and the correction coefficient is 0.8, then you actually deduct points: -200*0.002*0.8=-0.32 (approximately equal to 0, the ladder may be rounded off automatically), so you don't deduct points. Conversely, even if this low-scoring hero gets +200 hero points, the ladder score is +0. Why? Because the hero integral of your hero is far lower than the average of your top 30 heroes (step integral), your hero is at the low end of the weight coefficient of quasi-normal distribution, and the influence on the integral changes little.

So you asked again? So don't hit this low-scoring hero?

Wrong. When you make this low-scoring hero score higher than the top 30 average heroes, the hero weight coefficient will rise rapidly, which will significantly increase points. (In the process from low score to average score, points will be added slowly. The closer to the average score, the more obvious the integral will be. )

When all your heroes are in the upper middle of the normal distribution, your integral will eventually stabilize, that is to say, it is not obvious that you can add or subtract any hero (the change within 3 points). At this time, your ladder score is your real strength.

Why do many players show the fixed-point status prematurely?

That's because your heroic strength is also characterized by a stable normal distribution, and the results are mixed.

All in all,

Because of the existence of correction coefficient and weight coefficient, there are only two ways to play high-step integral:

1, only hit one hero, hit 1000 sets, won the game, and this hero was hit by your hero points to 2800. If the correction coefficient becomes 0.8 at this time, then your ladder integral is 2800*0.8=2240 (I once met a player who only played with a big ass. When the hero score was 3000, the ladder integral was more than 2300.

2. All heroes come on stage and hit the hero point of 2500. At this time, your correction factor is 0.9. Get the step integral =2250 (this is more in line with human nature, and the correction coefficient rises rapidly)

In fact, it is easier to get a complex mathematical formula by combining distribution function with probability modeling in mathematics.

I simplify it to a single number, which is easier to understand. I really don't know how these two coefficients change, but the function or point set of the correction coefficient should be convergent. The weight coefficient must conform to some probability distribution (similar to normal distribution).