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August 18, 2012 17:22
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Project Euler 問題27
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#!/usr/local/bin/octave -q | |
clear; | |
% n=0: f=b | |
% (∴) b は素数である必要がある | |
% n=1: f=1+a+b | |
% b は素数であることが確定している | |
% b=2の時: a は偶数 | |
% b=2以外: a は奇数 | |
% (∴) a は偶数である必要がある(0も偶数と見なして除外) | |
_b = primes(999); | |
m = 100; | |
n = [0:m]; | |
u = ones(1,m+1); | |
len_max = 0; | |
for(j=[1:1:length(_b)]) | |
b = _b(j); | |
if(b == 2) | |
_a = [-998:2:998]; | |
else | |
_a = [-999:2:999]; | |
end | |
for(i=[1:1:length(_a)]) | |
a = _a(i); | |
%calc | |
f = n.^2+a*n+b; | |
f = f.*(f > 0); | |
f2 = isprime(f); | |
len = find(xor(f2,u))(1)-1; | |
if(len > len_max) | |
printf("len = %d, a = %d, b = %d, Ans: a*b = %d\n", len, a, b, a*b); | |
len_max = len; | |
end | |
end | |
end | |
printf("(for check: last a = %d, b = %d\n", a, b); | |
% <出力> | |
% len = 71, a = -61, b = 971, Ans: a*b = -59231 |
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【TODO】aとbのループが遅い。行列計算でなんとか処理できないか考える。