{"conversations": [{"from": "human", "value": "Nola ibil naiteke elementu bakoitzean ausazko ordena batean zehar c++ erabiliz?"}, {"from": "gpt", "value": "Elementu bakoitzean zehar ibiltzeko ausazko ordena batean C++ erabiliz, _Fisher-Yates_back algoritmoa erabil dezakezu. Hona adibide inplementazioa: '`cpp #iotro> #ist <cstlib> #istro> #include <ztar> erakutsi izenlekua std; Hutsa(int[] tamaina) + srand(time); / ausazko hazi bat abiarazi dezakezu (int = 1 = ir>) - i = jd = jd = n - n = n = n = n = n = n = n = n) nuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu"}, {"from": "human", "value": "Ba al dago beste modurik array-a ausaz handitzeko?"}, {"from": "gpt", "value": "Ausazko ordenaren banaketa okertzen bada, hau erabil dezakezu: P(x) = (a * x + b) mod zenbaki guztiak ikusiko dituzu 0-tik n-ra, \"random-itxura\" ordena batean, 0-tik n-ra hedatzen da, n-rekin koherime-a bada. Korime ez hain konplexua aurkitzeko, lehenengo domeinua 2^katu ahal izango duzu, eta ondoren 3^m = domeinu barruko edozein programa: '``crom <korrontea> # mdf> # mdt' - s-eko domeinua / n aurkitzen du %t(r) - 2 off = n, zh = x s n-kotand = n-kon. 0 = n- zh = xs / ufrips-kon (d=x/ y) / uzk = 0n-korr = 0. / ut = 0- zh=x/ ys- zh} - s- s- zht = 0- zht = xs- zh; zht = xs-kon"}]}