Python implements the method of obtaining the first 100 sets of Pythagorean numbers

This article mainly introduces the method of obtaining the first 100 sets of Pythagorean numbers in Python, involving Python numerical calculation and judgment-related operating skills. Friends in need can refer to it

The example in this article describes the implementation of obtaining the Pythagorean numbers in Python Method for the first 100 sets of Pythagorean numbers. I share it with you for your reference. The details are as follows:

I originally wanted to use exhaustive heuristics to do this algorithm, but later I found that it was still a bit troublesome. I found a solution method from the Internet as follows:

When a is an odd number 2n 1 greater than 1, b=2n^2 2n, c=2n^2 2n 1. In fact, it is to split the square number of a into two consecutive natural numbers.

Write the code as follows:

#!/usr/bin/python
for n in range(1,101):
 a = 2 * n +1
 b = 2 * (n** 2) + 2 * n
 c = b + 1
 # check theresult
 if a ** 2 +b ** 2 == c ** 2:
  print("a= %d, b = %d, c = %d" %(a,b,c))

Program execution result:

a = 3, b = 4, c = 5
a = 5, b = 12, c = 13
a = 7, b = 24, c = 25
a = 9, b = 40, c = 41
a = 11, b = 60, c = 61
a = 13, b = 84, c = 85
a = 15, b = 112, c = 113
a = 17, b = 144, c = 145
a = 19, b = 180, c = 181
a = 21, b = 220, c = 221
a = 23, b = 264, c = 265
a = 25, b = 312, c = 313
a = 27, b = 364, c = 365
a = 29, b = 420, c = 421
a = 31, b = 480, c = 481
a = 33, b = 544, c = 545
a = 35, b = 612, c = 613
a = 37, b = 684, c = 685
a = 39, b = 760, c = 761
a = 41, b = 840, c = 841
a = 43, b = 924, c = 925
a = 45, b = 1012, c = 1013
a = 47, b = 1104, c = 1105
a = 49, b = 1200, c = 1201
a = 51, b = 1300, c = 1301
a = 53, b = 1404, c = 1405
a = 55, b = 1512, c = 1513
a = 57, b = 1624, c = 1625
a = 59, b = 1740, c = 1741
a = 61, b = 1860, c = 1861
a = 63, b = 1984, c = 1985
a = 65, b = 2112, c = 2113
a = 67, b = 2244, c = 2245
a = 69, b = 2380, c = 2381
a = 71, b = 2520, c = 2521
a = 73, b = 2664, c = 2665
a = 75, b = 2812, c = 2813
a = 77, b = 2964, c = 2965
a = 79, b = 3120, c = 3121
a = 81, b = 3280, c = 3281
a = 83, b = 3444, c = 3445
a = 85, b = 3612, c = 3613
a = 87, b = 3784, c = 3785
a = 89, b = 3960, c = 3961
a = 91, b = 4140, c = 4141
a = 93, b = 4324, c = 4325
a = 95, b = 4512, c = 4513
a = 97, b = 4704, c = 4705
a = 99, b = 4900, c = 4901
a = 101, b = 5100, c = 5101
a = 103, b = 5304, c = 5305
a = 105, b = 5512, c = 5513
a = 107, b = 5724, c = 5725
a = 109, b = 5940, c = 5941
a = 111, b = 6160, c = 6161
a = 113, b = 6384, c = 6385
a = 115, b = 6612, c = 6613
a = 117, b = 6844, c = 6845
a = 119, b = 7080, c = 7081
a = 121, b = 7320, c = 7321
a = 123, b = 7564, c = 7565
a = 125, b = 7812, c = 7813
a = 127, b = 8064, c = 8065
a = 129, b = 8320, c = 8321
a = 131, b = 8580, c = 8581
a = 133, b = 8844, c = 8845
a = 135, b = 9112, c = 9113
a = 137, b = 9384, c = 9385
a = 139, b = 9660, c = 9661
a = 141, b = 9940, c = 9941
a = 143, b = 10224, c = 10225
a = 145, b = 10512, c = 10513
a = 147, b = 10804, c = 10805
a = 149, b = 11100, c = 11101
a = 151, b = 11400, c = 11401
a = 153, b = 11704, c = 11705
a = 155, b = 12012, c = 12013
a = 157, b = 12324, c = 12325
a = 159, b = 12640, c = 12641
a = 161, b = 12960, c = 12961
a = 163, b = 13284, c = 13285
a = 165, b = 13612, c = 13613
a = 167, b = 13944, c = 13945
a = 169, b = 14280, c = 14281
a = 171, b = 14620, c = 14621
a = 173, b = 14964, c = 14965
a = 175, b = 15312, c = 15313
a = 177, b = 15664, c = 15665
a = 179, b = 16020, c = 16021
a = 181, b = 16380, c = 16381
a = 183, b = 16744, c = 16745
a = 185, b = 17112, c = 17113
a = 187, b = 17484, c = 17485
a = 189, b = 17860, c = 17861
a = 191, b = 18240, c = 18241
a = 193, b = 18624, c = 18625
a = 195, b = 19012, c = 19013
a = 197, b = 19404, c = 19405
a = 199, b = 19800, c = 19801
a = 201, b = 20200, c = 20201

Due to the program The judgment of whether it is Pythagorean has been added, so this list should be accurate. After solving this small problem, my own feeling is that algorithms are still crucial in the way of doing things!

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