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) n$ t+ I2 A, b6 Finclude<stdio.h>///////////////// 适合第一题 #include<math.h> #define N 1000 : {" r! o& @5 e- W
double f(double x) { return pow(x,-2);}
8 h+ s( g5 O6 J- u0 X- H& ^" B7 Udouble 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;}
) d- J/ I8 W0 M9 v M. T2 {. D. l: k 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);
, j. f4 ^- Q' x3 b: Z! F 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;
4 Z; `: m) a$ E 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; }
+ s" X& ]% o7 t. A/ f$ C* v# K# _$ nvoid main() { int i,n2; / x# f3 s& a) @4 L
double a,b,h=0.0,e=0.0,len=0.0,T=0.0; 5 }( z+ Y4 U0 M9 P( o" F+ L
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);
" b1 O+ b3 k3 a/ M8 Z- q5 R" ^ 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); ) U! Y1 D7 U+ P1 y# {
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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
K: a3 f. L. c [" A a; M每排单号数字表示DVD编号,偶号数字为起满意度 9 T- @" r$ R9 U; ^( `' n
%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; 3 `+ M% A0 @0 U( P% B, x
%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 & F! N; {9 K) z' `( ]& U/ u/ m
%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
& _! a4 Y. Y" f% X6 S3 C%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;
8 b! g2 h3 ^$ U: l! X3 W$ w%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; ' \: T+ v- [* t2 o8 Y
%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 ) W9 _9 V0 N) a- i5 L; {+ Z
%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
% q! O5 r7 I5 P- Y6 w, R2 n%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
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