2005年B题DVD租赁完整答案
<P>6250</P><P>include<stdio.h>///////////////// 适合第一题<BR>#include<math.h><BR>#define N 1000</P>
<P>double f(double x)<BR>{ return pow(x,-2);}</P>
<P>double zhongzhi(double a,double h,int n1)<BR>{double m=0.0;<BR> int k;<BR> for(k=0;k<=n1;k++)<BR> m=m+f((k*h+a+(k+1)*h+a)/2);<BR>return m;}</P>
<P><BR>double pow1(int k,int r)<BR>{ <BR>int f=1,i;<BR>for(i=0;i<r;i++)<BR>f*=k;<BR>return f;<BR>}<BR> <BR>double longbg(double a,double b,double e )<BR>{ long int n=1;<BR>int i,j,p,q;<BR>double h=0.0,m1=0.0,g={0};<BR> h=b-a; <BR>g=h/2*(f(a)+f(b)); <BR>for(i=1;i<N;i++)<BR> { g=(double)1/2*g+h/2*zhongzhi(a,h,n-1); </P>
<P> <BR> for(j=1;j<=i;j++)<BR> <BR> g=pow1(4,j)/(pow1(4,j)-1)*g-1/(pow1(4,j)-1)*g;<BR> <BR> <BR> <BR> n*=2;<BR> <BR> h=(double)(b-a)/n;<BR> </P>
<P> <BR> <BR> if(fabs(g-g)<e) break; <BR> p=i;q=j;<BR> <BR> }<BR>/* printf("龙贝格矩阵为:\n");<BR> for(i=0;i<=p;i++){<BR> printf("\n");<BR> for(j=0;j<=i;j++)<BR> printf(" %.7lf",g);}<BR> <BR> printf("\n满意 积分的值为 :%.7lf\n",g);*/<BR>m1=g;<BR> return m1;<BR>}</P>
<P>void main()<BR>{<BR>int i,n2;</P>
<P><BR>double a,b,h=0.0,e=0.0,len=0.0,T=0.0;</P>
<P>printf("请输入所满意积分的下界和误差极限:A,e:");<BR>scanf("%lf%lf",&a,&e);<BR>printf("输入DVD 的个数N2和 类别 len:\n");<BR>scanf("%d %lf",&n2,&len);<BR> printf("所输入的日期 初始点为=%lf 分周期 为 = %d",a,n2); </P>
<P> <BR> b=a+len;<BR>for(i=1;i<=n2;i++)<BR>{T=T+longbg(a,b,e);<BR> a=b;<BR> b=a+len;<BR>if(f(b)<e)break;<BR>}<BR>printf(" 所求的满意度常积分为: %lf",T);</P>
<P>8 1 82 2 98 3 1<BR>6 1 44 2 62 4 2<BR>80 1 50 2 32 4 3<BR>7 1 18 2 41 3 4<BR>66 1 68 2 11 3 5<BR>19 1 53 2 66 4 6<BR>81 1 8 2 26 3 7<BR>71 1 31 4 35 5 8<BR>53 1 100 2 78 3 9<BR>60 1 55 2 85 3 10<BR>59 1 63 2 19 3 11<BR>31 1 2 2 7 3 12<BR>96 1 78 2 21 3 13<BR>52 1 23 2 89 6 14<BR>13 1 85 3 52 4 15<BR>84 1 97 2 10 4 16<BR>67 1 47 2 51 3 17<BR>41 1 60 2 78 3 18<BR>84 1 86 2 66 4 19<BR>45 1 89 2 61 3 20<BR>53 1 45 2 2 4 21<BR>57 1 55 2 38 3 22<BR>95 1 29 2 81 3 23<BR>76 1 41 2 37 4 24<BR>9 1 69 2 81 4 25<BR>22 1 68 2 95 3 26<BR>58 1 22 3 50 4 27<BR>8 1 34 2 47 6 28<BR>55 1 30 2 44 3 29<BR>62 1 37 2 70 4 30</P>
<P>每排单号数字表示DVD编号,偶号数字为起满意度</P>
<P>%dingdan(1000),dvd(100)<BR>tt(1000)=0; %顾客当前可租次数,初始值为随机分布<BR>finish(1000)=0; %当前已租赁,1为第一次租,2为第二次租<BR>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>xuqiu(1000)(8)=0; %需求矩阵<BR>zulin(1000)(6)=0;</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>for d_i=1:1000<BR> for d_j=1:100<BR> if dingdan(d_i)(d_j) > 0 & dingdan(d_i)(d_j) < 9<BR> xuqiu(d_i)(dingdan(d_i)(d_j)) = d_j;<BR> end<BR> end<BR>end</P>
<P>%产生两个随机分布<BR>tt_temp=rand(1,1000);<BR>time_temp=rand(1,1000);<BR>for i=1:1000<BR> if tt_temp(i) < 0.6<BR> tt(i) = 2;<BR> else <BR> tt(i) = 1;<BR> end<BR> if tt(i) == 2<BR> time(i) = floor(time_temp(i)*12+3);<BR> else<BR> time(i) = floor(time_temp(i)*27+3);<BR> end<BR>end </P>
<P>%步长为一天,计算30天<BR>for i = 1:30<BR> %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> for j=1:1000<BR> if finish(j) == 1 | finish(j) == 2<BR> time(j) = time(j) - 1; <BR> if time(j) == 0 <BR> if finish(j) == 1; %如果为第一次租,则归还后dvd数量增加<BR> dvd(zulin(j)(1)) = dvd(zulin(j)(1)) + 1;<BR> dvd(zulin(j)(2)) = dvd(zulin(j)(2)) + 1;<BR> dvd(zulin(j)(3)) = dvd(zulin(j)(3)) + 1;<BR> end<BR> if finish(j) == 2; %如果为第二次租,则归还后dvd数量增加<BR> dvd(zulin(j)(4)) = dvd(zulin(j)(4)) + 1;<BR> dvd(zulin(j)(5)) = dvd(zulin(j)(5)) + 1;<BR> dvd(zulin(j)(6)) = dvd(zulin(j)(6)) + 1;<BR> end<BR> end<BR> end<BR> end<BR> %遍历所有顾客,如果需要租赁,则进行分发处理<BR> for k=1:1000<BR> if tt(k) ==2 %能租两次的第一次租的处理<BR> num_flag = 0;<BR> for dvdnum=1:6<BR> if dvd(xiuqiu(k)(dvdnum)) > 0<BR> num_flag = num_flag + 1;<BR> end<BR> end<BR> if num_flag > 2 %能租到3张碟,则出租最靠前满意度的三张碟<BR> dvd_2zu_i=1;<BR> dvd_2zu_3=1;<BR> while dvd_2zu_3 <= 3 %决定哪三张碟<BR> if dvd(xiuqiu(k)(vd_2zu_i)) > 0<BR> dvd(xiuqiu(k)(vd_2zu_i)) = dvd(xiuqiu(k)(vd_2zu_i)) -1;<BR> zulin(k)(dvd_2zu_3) = xiuqiu(k)(vd_2zu_i);<BR> dvd_2zu_3 = dvd_2zu_3 + 1;<BR> end<BR> dvd_2zu_i = dvd_2zu_i + 1;<BR> end<BR> tt(k) = 1;<BR> finish(k) = 2;</P>
<P>%dingdan(1000),dvd(100)<BR>tt(1000)=0; %顾客当前可租次数,初始值为随机分布<BR>finish(1000)=0; %当前已租赁,1为第一次租,2为第二次租<BR>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>xuqiu(1000)(8)=0; %需求矩阵<BR>zulin(1000)(6)=0;</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>for d_i=1:1000<BR> for d_j=1:100<BR> if dingdan(d_i)(d_j) > 0 & dingdan(d_i)(d_j) < 9<BR> xuqiu(d_i)(dingdan(d_i)(d_j)) = d_j;<BR> end<BR> end<BR>end</P>
<P>%产生两个随机分布<BR>tt_temp=rand(1,1000);<BR>time_temp=rand(1,1000);<BR>for i=1:1000<BR> if tt_temp(i) < 0.6<BR> tt(i) = 2;<BR> else <BR> tt(i) = 1;<BR> end<BR> if tt(i) == 2<BR> time(i) = floor(time_temp(i)*12+3);<BR> else<BR> time(i) = floor(time_temp(i)*27+3);<BR> end<BR>end </P>
<P>%步长为一天,计算30天<BR>for i = 1:30<BR> %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> for j=1:1000<BR> if finish(j) == 1 | finish(j) == 2<BR> time(j) = time(j) - 1; <BR> if time(j) == 0 <BR> if finish(j) == 1; %如果为第一次租,则归还后dvd数量增加<BR> dvd(zulin(j)(1)) = dvd(zulin(j)(1)) + 1;<BR> dvd(zulin(j)(2)) = dvd(zulin(j)(2)) + 1;<BR> dvd(zulin(j)(3)) = dvd(zulin(j)(3)) + 1;<BR> end<BR> if finish(j) == 2; %如果为第二次租,则归还后dvd数量增加<BR> dvd(zulin(j)(4)) = dvd(zulin(j)(4)) + 1;<BR> dvd(zulin(j)(5)) = dvd(zulin(j)(5)) + 1;<BR> dvd(zulin(j)(6)) = dvd(zulin(j)(6)) + 1;<BR> end<BR> end<BR> end<BR> end<BR> %遍历所有顾客,如果需要租赁,则进行分发处理<BR> for k=1:1000<BR> if tt(k) ==2 %能租两次的第一次租的处理<BR> num_flag = 0;<BR> for dvdnum=1:6<BR> if dvd(xiuqiu(k)(dvdnum)) > 0<BR> num_flag = num_flag + 1;<BR> end<BR> end<BR> if num_flag > 2 %能租到3张碟,则出租最靠前满意度的三张碟<BR> dvd_2zu_i=1;<BR> dvd_2zu_3=1;<BR> while dvd_2zu_3 <= 3 %决定哪三张碟<BR> if dvd(xiuqiu(k)(vd_2zu_i)) > 0<BR> dvd(xiuqiu(k)(vd_2zu_i)) = dvd(xiuqiu(k)(vd_2zu_i)) -1;<BR> zulin(k)(dvd_2zu_3) = xiuqiu(k)(vd_2zu_i);<BR> dvd_2zu_3 = dvd_2zu_3 + 1;<BR> end<BR> dvd_2zu_i = dvd_2zu_i + 1;<BR> end</P>
<P>31 46 37 48 31 38 40 43 45 35<BR>39 41 38 41 37 48 38 36 41 48<BR>44 39 45 32 39 41 36 29 35 51<BR>39 45 41 41 49 45 31 40 39 38<BR>63 45 36 45 44 35 42 35 42 44<BR>50 36 43 36 41 42 41 38 44 46<BR>37 41 42 44 42 41 40 45 43 40<BR>46 44 35 41 38 34 30 40 41 38<BR>39 28 32 29 43 31 44 34 34 37<BR>49 38 34 33 50 34 46 42 28 45</P> 什么东西吗 怎么回事啊 什么垃圾网站 没钱会死你啊 shi <P>你妈妈的是不是不想看今天的月亮了</P>
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