Maple编程解题的例子

maplelab

题目如图所示:


以下为有限解决的代码:

生成满足第一个和条件的数:

1. restart; sumtest := proc (n) local P, sumP, subP, r, dataP, i, j; P := proc (n) (1/2)*n*(3*n-1) end proc; sumP := proc (a, b) P(a)+P(b) end proc; subP := proc (a, b) P(a)-P(b) end proc; r := proc (a, b) abs(subP(a, b)) end proc; dataP := proc (n) [seq(P(i), i = 1 .. n)] end proc; for i to nops(dataP(n)) do for j to nops(dataP(n)) do if member(sumP(i, j), dataP(n)) then print([[P(i), P(j), sumP(i, j)]]) end if end do end do end proc

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这样的数容易找到.

生成满足第二个差条件的数:

1. subtest := proc (n) local P, sumP, subP, r, dataP, i, j; P := proc (n) (1/2)*n*(3*n-1) end proc; sumP := proc (a, b) P(a)+P(b) end proc; subP := proc (a, b) P(a)-P(b) end proc; r := proc (a, b) abs(subP(a, b)) end proc; dataP := proc (n) [seq(P(i), i = 1 .. n)] end proc; for i to nops(dataP(n)) do for j to nops(dataP(n)) do if member(subP(i, j), dataP(n)) then print([[P(i), P(j), subP(i, j)]]) end if end do end do end proc

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这样的数也容易找到.

生成同时满足和条件和差条件的数:

1. test := proc (n) local P, sumP, subP, r, dataP, i, j; P := proc (n) (1/2)*n*(3*n-1) end proc; sumP := proc (a, b) P(a)+P(b) end proc; subP := proc (a, b) P(a)-P(b) end proc; r := proc (a, b) abs(subP(a, b)) end proc; dataP := proc (n) [seq(P(i), i = 1 .. n)] end proc; for i to nops(dataP(n)) do for j to nops(dataP(n)) do if member(sumP(i, j), dataP(n)) and member(subP(i, j), dataP(n)) then print([P(i), P(j), [sumP(i, j), subP(i, j)]]) end if end do end do end proc

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由于在很大范围内没有发现能同时满足前两个条件的数,倒是可以怀疑是否存在同时满足三个条件的数.