2005年B题DVD租赁完整答案

123456

6250

/ U$ T$ D/ P. m8 F, ?& t2 d

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

' a0 V1 \6 n; `' ?6 V

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

2 x5 S2 G$ r4 F, t+ y

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;}

) f* k$ L* p% r! _6 o1 I

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);

7 a& @ K- {- ]2 V. l/ Q; \

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;

9 B; n0 s. O" Y Q

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);

# K' |. T ^- y' q3 t$ S% t* P' ] + v) ~: o9 n3 Z ) p1 q6 k5 I( L3 V. w

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

. N: V: z9 E$ d1 G/ p3 h- ~

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;

# }1 ]* `7 X2 I1 }

%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

3 q) g. X% @ P

%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

%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;

: }, O* l( W+ L$ C+ S1 a( I) ]+ h( 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;

%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

6 O/ g/ v) x( U$ D+ r

%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

6 @ v B- @: E+ ~$ ~

%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

3 [$ W% \3 J: Y

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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