Perfect Cubes |
尚未結案
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writeman
初階會員 ![]() ![]() 發表:31 回覆:34 積分:28 註冊:2004-02-17 發送簡訊給我 |
For hundreds of years Fermat's Last Theorem,
which stated simply that for n > 2 there exist no integers a, b, c > 1
such that a^n = b^n c^n, has remained elusively unproven.
(A recent proof is believed to be correct,
though it is still undergoing scrutiny.)
It is possible, however, to find integers greater than 1 that satisfy
the ``perfect cube'' equation a^3=b^3 c^3 d^3
(e.g. a quick calculation will show that the equation
12^3 = 6^3 8^3 10^3is indeed true).
This problem requires that you write a program to find all sets of numbers
{a, b, c, d} which satisfy this equation for .
The first part of the output is shown here:
Cube = 6, Triple = (3,4,5)
Cube = 12, Triple = (6,8,10)
Cube = 18, Triple = (2,12,16)
Cube = 18, Triple = (9,12,15)
Cube = 19, Triple = (3,10,18)
Cube = 20, Triple = (7,14,17)
Cube = 24, Triple = (12,16,20)
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無故障
一般會員 ![]() ![]() 發表:17 回覆:69 積分:17 註冊:2004-03-11 發送簡訊給我 |
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無故障
一般會員 ![]() ![]() 發表:17 回覆:69 積分:17 註冊:2004-03-11 發送簡訊給我 |
供參考
這個好像是作業....
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