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dp_subd
dp_ht(dpoly1)/dp_ht(dpoly2)
を求める. 結果の係数は 1
である.
[162] dp_subd(<<1,2,3,4,5>>,<<1,1,2,3,4>>); (1)*<<0,1,1,1,1>>
dp_red
, dp_red_mod
.
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