%dingdan(1000),dvd(100) <br>
<>
<br>
<p>tt(1000)=0; %顾客当前可租次数,初始值为随机分布 <br>
<>
<p>
<p>finish(1000)=0; %当前已租赁,1为第一次租,2为第二次租 <br>
<>
<p>
<p>time(1000)=0; %顾客保存周期几天,租一次为3~30天,租两次为3-15天,随机分布 <br>
<P>
<p>
<p>xuqiu(1000)(8)=0; %需求矩阵 <br>
<P>
<p>
<p>zulin(1000)(6)=0; <br>
<P>
<p>
<p>
<P>
<p>
<p>
<P>
<p>
<p>
<P>
<p>
<p>%转换订单矩阵为需求矩阵,xuqiu(d_i)(dingdan(d_i)(d_j))值为喜好程度为d_j的dvd编号 <br>
<P>
<p>
<p>for d_i=1:1000 <br>
<P>
<p>
<p> for d_j=1:100 <br>
<P>
<p>
<p> if dingdan(d_i)(d_j) > 0 & dingdan(d_i)(d_j) < 9 <br>
<P>
<p>
<p> xuqiu(d_i)(dingdan(d_i)(d_j)) = d_j; <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p>end <br>
<P>
<p>
<p>
<P>
<p>
<p>
<P>
<p>
<p>
<P>
<p>
<p>%产生两个随机分布 <br>
<P>
<p>
<p>tt_temp=rand(1,1000); <br>
<P>
<p>
<p>time_temp=rand(1,1000); <br>
<P>
<p>
<p>for i=1:1000 <br>
<P>
<p>
<p> if tt_temp(i) < 0.6 <br>
<P>
<p>
<p> tt(i) = 2; <br>
<P>
<p>
<p> else <br>
<P>
<p>
<p> tt(i) = 1; <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> if tt(i) == 2 <br>
<P>
<p>
<p> time(i) = floor(time_temp(i)*12+3); <br>
<P>
<p>
<p> else <br>
<P>
<p>
<p> time(i) = floor(time_temp(i)*27+3); <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p>end <br>
<P>
<p>
<p>
<P>
<p>
<p>
<P>
<p>
<p>
<P>
<p>
<p>%步长为一天,计算30天 <br>
<P>
<p>
<p>for i = 1:30 <br>
<P>
<p>
<p> %遍历所有已租赁顾客,是否有到保存周期,归还,如果有第二次借则再借 <br>
<P>
<p>
<p> for j=1:1000 <br>
<P>
<p>
<p> if finish(j) == 1 | finish(j) == 2 <br>
<P>
<p>
<p> time(j) = time(j) - 1; <br>
<P>
<p>
<p> if time(j) == 0 <br>
<P>
<p>
<p> if finish(j) == 1; %如果为第一次租,则归还后dvd数量增加 <br>
<P>
<p>
<p> dvd(zulin(j)(1)) = dvd(zulin(j)(1)) + 1; <br>
<P>
<p>
<p> dvd(zulin(j)(2)) = dvd(zulin(j)(2)) + 1; <br>
<P>
<p>
<p> dvd(zulin(j)(3)) = dvd(zulin(j)(3)) + 1; <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> if finish(j) == 2; %如果为第二次租,则归还后dvd数量增加 <br>
<P>
<p>
<p> dvd(zulin(j)(4)) = dvd(zulin(j)(4)) + 1; <br>
<P>
<p>
<p> dvd(zulin(j)(5)) = dvd(zulin(j)(5)) + 1; <br>
<P>
<p>
<p> dvd(zulin(j)(6)) = dvd(zulin(j)(6)) + 1; <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> %遍历所有顾客,如果需要租赁,则进行分发处理 <br>
<P>
<p>
<p> for k=1:1000 <br>
<P>
<p>
<p> if tt(k) ==2 %能租两次的第一次租的处理 <br>
<P>
<p>
<p> num_flag = 0; <br>
<P>
<p>
<p> for dvdnum=1:6 <br>
<P>
<p>
<p> if dvd(xiuqiu(k)(dvdnum)) > 0 <br>
<P>
<p>
<p> num_flag = num_flag + 1; <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> if num_flag > 2 %能租到3张碟,则出租最靠前满意度的三张碟 <br>
<P>
<p>
<p> dvd_2zu_i=1; <br>
<P>
<p>
<p> dvd_2zu_3=1; <br>
<P>
<p>
<p> while dvd_2zu_3 <= 3 %决定哪三张碟 <br>
<P>
<p>
<p> if dvd(xiuqiu(k)(vd_2zu_i)) > 0 <br>
<P>
<p>
<p> dvd(xiuqiu(k)(vd_2zu_i)) = dvd(xiuqiu(k)(vd_2zu_i)) -1; <br>
<P>
<p>
<p> zulin(k)(dvd_2zu_3) = xiuqiu(k)(vd_2zu_i); <br>
<P>
<p>
<p> dvd_2zu_3 = dvd_2zu_3 + 1; <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> dvd_2zu_i = dvd_2zu_i + 1; <br>
<P>
<p>
<p> end <br>
<P>
<p>
<p> <br>
<P>这有段D题的程序```大家看看`不懂再问!!!!!看看有什么问题不??</P>
[此贴子已经被作者于2005-9-17 11:39:17编辑过]
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