如何生成与rand(1..117)结果偏差±20且限1-117的随机数？

Got it, let's walk through how to generate that second random number exactly as you need it.

First, let's recap your requirements clearly:

• You start with a random number from rand(1..117) (you mentioned getting 80 as an example)
• The second number needs to be within ±20 of the first one
• It also has to stay strictly between 1 and 117 (no negatives, no numbers over 117)

Step-by-Step Solution #

The key here is to calculate a valid range for the second number that respects both the ±20 constraint and the 1-117 bounds. Here's how to do it in Ruby (since you used Ruby's rand syntax):

# First, get your initial random number (we'll use 80 as your example)
first_num = 80 # or rand(1..117) to generate it dynamically

# Calculate the lower bound: don't go below 1
lower_bound = [first_num - 20, 1].max
# Calculate the upper bound: don't go above 117
upper_bound = [first_num + 20, 117].min

# Generate the second random number within this safe range
second_num = rand(lower_bound..upper_bound)

How This Works #

• When your first number is 80: lower_bound becomes 80-20=60 (which is above 1, so we keep it), and upper_bound becomes 80+20=100 (which is below 117, so we keep it). So your second number will be between 60 and 100—all values here are within ±20 of 80 and stay in the 1-117 range.
• If your first number was closer to a boundary (like 15): lower_bound would be 1 (since 15-20=-5, we cap it at 1), and upper_bound is 35. So you get a number between 1 and 35.
• If your first number was 110: lower_bound is 90, upper_bound is 117 (since 110+20=130 exceeds 117, we cap it). So you get a number between 90 and 117.

Quick Note on Your Examples #

A heads-up: some of the numbers you listed (like 50, 35, 108) are actually outside the ±20 range from 80. For example, 80-50=30, which is more than 20. Valid examples would be 64, 79, 99, 88—all of these are within 60-100, so they fit both constraints perfectly.

内容的提问来源于stack exchange，提问作者chris