Python实现获取前实现获取前100组勾股数的方法示例组勾股数的方法示例
本文实例讲述了Python实现获取前100组勾股数的方法。分享给大家供大家参考，具体如下：
本来想采用穷举试探的方式来做这个算法，后来发现还是有点麻烦。从网络上找来了一种求解方法如下：
当a为大于1的奇数2n+1时，b=2n^2+2n,c=2n^2+2n+1。实际上就是把a的平方数拆成两个连续自然数。
编写代码如下：
#!/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))
程序执行结果：
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
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