2005年B题DVD租赁完整答案

123456

6250

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include<stdio.h>///////////////// 适合第一题
#include<math.h>
#define N 1000

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double f(double x)
{ return pow(x,-2);}

double zhongzhi(double a,double h,int n1)
{double m=0.0;
int k;
for(k=0;k<=n1;k++)
m=m+f((k*h+a+(k+1)*h+a)/2);
return m;}

3 C( q, i! x- s5 r7 N' \6 G

double pow1(int k,int r)
{
int f=1,i;
for(i=0;i<r;i++)
f*=k;
return f;
}

double longbg(double a,double b,double e )
{ long int n=1;
int i,j,p,q;
double h=0.0,m1=0.0,g[N][N]={0};
h=b-a;
g[0][0]=h/2*(f(a)+f(b));
for(i=1;i<N;i++)
{ g[0]=(double)1/2*g[i-1][0]+h/2*zhongzhi(a,h,n-1);

for(j=1;j<=i;j++)

g[j]=pow1(4,j)/(pow1(4,j)-1)*g[j-1]-1/(pow1(4,j)-1)*g[i-1][j-1];



n*=2;

h=(double)(b-a)/n;

% R# \8 {' ^/ N5 r& \9 F" E1 s

if(fabs(g-g[i-1][i-1])<e) break;
p=i;q=j;

}
/* printf("龙贝格矩阵为：\n");
for(i=0;i<=p;i++){
printf("\n");
for(j=0;j<=i;j++)
printf(" %.7lf",g[j]);}

printf("\n满意 积分的值为 ：%.7lf\n",g[p][q-1]);*/
m1=g[p][q-1];
return m1;
}

void main()
{
int i,n2;

double a,b,h=0.0,e=0.0,len=0.0,T=0.0;

printf("请输入所满意积分的下界和误差极限：A,e：");
scanf("%lf%lf",&a,&e);
printf("输入DVD 的个数N2和 类别 len:\n");
scanf("%d %lf",&n2,&len);
printf("所输入的日期 初始点为=%lf 分周期 为 = %d",a,n2);

b=a+len;
for(i=1;i<=n2;i++)
{T=T+longbg(a,b,e);
a=b;
b=a+len;
if(f(b)<e)break;
}
printf(" 所求的满意度常积分为： %lf",T);

4 v, j- ~) E: k0 l' r9 I$ n ! S1 \1 U2 v' i% Y: ~, C! d6 {

8 1 82 2 98 3 1
6 1 44 2 62 4 2
80 1 50 2 32 4 3
7 1 18 2 41 3 4
66 1 68 2 11 3 5
19 1 53 2 66 4 6
81 1 8 2 26 3 7
71 1 31 4 35 5 8
53 1 100 2 78 3 9
60 1 55 2 85 3 10
59 1 63 2 19 3 11
31 1 2 2 7 3 12
96 1 78 2 21 3 13
52 1 23 2 89 6 14
13 1 85 3 52 4 15
84 1 97 2 10 4 16
67 1 47 2 51 3 17
41 1 60 2 78 3 18
84 1 86 2 66 4 19
45 1 89 2 61 3 20
53 1 45 2 2 4 21
57 1 55 2 38 3 22
95 1 29 2 81 3 23
76 1 41 2 37 4 24
9 1 69 2 81 4 25
22 1 68 2 95 3 26
58 1 22 3 50 4 27
8 1 34 2 47 6 28
55 1 30 2 44 3 29
62 1 37 2 70 4 30

每排单号数字表示DVD编号，偶号数字为起满意度

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%dingdan(1000),dvd(100)
tt(1000)=0; %顾客当前可租次数，初始值为随机分布
finish(1000)=0; %当前已租赁，1为第一次租，2为第二次租
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
xuqiu(1000)(8)=0; %需求矩阵
zulin(1000)(6)=0;

%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
for d_i=1:1000
for d_j=1:100
if dingdan(d_i)(d_j) > 0 & dingdan(d_i)(d_j) < 9
xuqiu(d_i)(dingdan(d_i)(d_j)) = d_j;
end
end
end

%产生两个随机分布
tt_temp=rand(1,1000);
time_temp=rand(1,1000);
for i=1:1000
if tt_temp(i) < 0.6
tt(i) = 2;
else
tt(i) = 1;
end
if tt(i) == 2
time(i) = floor(time_temp(i)*12+3);
else
time(i) = floor(time_temp(i)*27+3);
end
end

0 `0 Z4 H; A' d. I' C6 H$ l

%步长为一天，计算30天
for i = 1:30
%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
for j=1:1000
if finish(j) == 1 | finish(j) == 2
time(j) = time(j) - 1;
if time(j) == 0
if finish(j) == 1; %如果为第一次租，则归还后dvd数量增加
dvd(zulin(j)(1)) = dvd(zulin(j)(1)) + 1;
dvd(zulin(j)(2)) = dvd(zulin(j)(2)) + 1;
dvd(zulin(j)(3)) = dvd(zulin(j)(3)) + 1;
end
if finish(j) == 2; %如果为第二次租，则归还后dvd数量增加
dvd(zulin(j)(4)) = dvd(zulin(j)(4)) + 1;
dvd(zulin(j)(5)) = dvd(zulin(j)(5)) + 1;
dvd(zulin(j)(6)) = dvd(zulin(j)(6)) + 1;
end
end
end
end
%遍历所有顾客，如果需要租赁，则进行分发处理
for k=1:1000
if tt(k) ==2 %能租两次的第一次租的处理
num_flag = 0;
for dvdnum=1:6
if dvd(xiuqiu(k)(dvdnum)) > 0
num_flag = num_flag + 1;
end
end
if num_flag > 2 %能租到3张碟，则出租最靠前满意度的三张碟
dvd_2zu_i=1;
dvd_2zu_3=1;
while dvd_2zu_3 <= 3 %决定哪三张碟
if dvd(xiuqiu(k)(vd_2zu_i)) > 0
dvd(xiuqiu(k)(vd_2zu_i)) = dvd(xiuqiu(k)(vd_2zu_i)) -1;
zulin(k)(dvd_2zu_3) = xiuqiu(k)(vd_2zu_i);
dvd_2zu_3 = dvd_2zu_3 + 1;
end
dvd_2zu_i = dvd_2zu_i + 1;
end
tt(k) = 1;
finish(k) = 2;

0 H, X3 y# e0 y3 a% C; d

%dingdan(1000),dvd(100)
tt(1000)=0; %顾客当前可租次数，初始值为随机分布
finish(1000)=0; %当前已租赁，1为第一次租，2为第二次租
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
xuqiu(1000)(8)=0; %需求矩阵
zulin(1000)(6)=0;

# d3 B+ H- b! \ V

%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
for d_i=1:1000
for d_j=1:100
if dingdan(d_i)(d_j) > 0 & dingdan(d_i)(d_j) < 9
xuqiu(d_i)(dingdan(d_i)(d_j)) = d_j;
end
end
end

%产生两个随机分布
tt_temp=rand(1,1000);
time_temp=rand(1,1000);
for i=1:1000
if tt_temp(i) < 0.6
tt(i) = 2;
else
tt(i) = 1;
end
if tt(i) == 2
time(i) = floor(time_temp(i)*12+3);
else
time(i) = floor(time_temp(i)*27+3);
end
end

]' H1 [# V q+ f4 M9 l# D S

%步长为一天，计算30天
for i = 1:30
%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
for j=1:1000
if finish(j) == 1 | finish(j) == 2
time(j) = time(j) - 1;
if time(j) == 0
if finish(j) == 1; %如果为第一次租，则归还后dvd数量增加
dvd(zulin(j)(1)) = dvd(zulin(j)(1)) + 1;
dvd(zulin(j)(2)) = dvd(zulin(j)(2)) + 1;
dvd(zulin(j)(3)) = dvd(zulin(j)(3)) + 1;
end
if finish(j) == 2; %如果为第二次租，则归还后dvd数量增加
dvd(zulin(j)(4)) = dvd(zulin(j)(4)) + 1;
dvd(zulin(j)(5)) = dvd(zulin(j)(5)) + 1;
dvd(zulin(j)(6)) = dvd(zulin(j)(6)) + 1;
end
end
end
end
%遍历所有顾客，如果需要租赁，则进行分发处理
for k=1:1000
if tt(k) ==2 %能租两次的第一次租的处理
num_flag = 0;
for dvdnum=1:6
if dvd(xiuqiu(k)(dvdnum)) > 0
num_flag = num_flag + 1;
end
end
if num_flag > 2 %能租到3张碟，则出租最靠前满意度的三张碟
dvd_2zu_i=1;
dvd_2zu_3=1;
while dvd_2zu_3 <= 3 %决定哪三张碟
if dvd(xiuqiu(k)(vd_2zu_i)) > 0
dvd(xiuqiu(k)(vd_2zu_i)) = dvd(xiuqiu(k)(vd_2zu_i)) -1;
zulin(k)(dvd_2zu_3) = xiuqiu(k)(vd_2zu_i);
dvd_2zu_3 = dvd_2zu_3 + 1;
end
dvd_2zu_i = dvd_2zu_i + 1;
end

31 46 37 48 31 38 40 43 45 35
39 41 38 41 37 48 38 36 41 48
44 39 45 32 39 41 36 29 35 51
39 45 41 41 49 45 31 40 39 38
63 45 36 45 44 35 42 35 42 44
50 36 43 36 41 42 41 38 44 46
37 41 42 44 42 41 40 45 43 40
46 44 35 41 38 34 30 40 41 38
39 28 32 29 43 31 44 34 34 37
49 38 34 33 50 34 46 42 28 45

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