2003 B题 露天矿卡车 卓金武程序中A b 怎么来的？？？？？？？

c=[5.265.194.2142.952.742.461.90.641.271.90.991.91.131.272.251.482.043.093.514.423.863.723.162.252.810.781.621.270.55.895.615.614.563.513.652.462.461.060.570.641.761.271.832.742.64.213.725.056.1];
A=[0.01540.01540.01540.01540.01540.01540.01540.01540.01540.01540000000000000000000000000000000000000000
-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.01540000000000000000000000000000000000000000
00000000000.01540.01540.01540.01540.01540.01540.01540.01540.01540.0154000000000000000000000000000000
0000000000-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154000000000000000000000000000000
000000000000000000000.01540.01540.01540.01540.01540.01540.01540.01540.01540.015400000000000000000000
00000000000000000000-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.0154-0.015400000000000000000000
-0.005-0.025-0.0150.0150.0050.0250.0150.0050.0250.0050000000000000000000000000000000000000000
-0.01040.0096-0.0004-0.0304-0.0204-0.0404-0.0304-0.0204-0.0404-0.02040000000000000000000000000000000000000000
0000000000-0.005-0.025-0.0150.0150.0050.0250.0150.0050.0250.005000000000000000000000000000000
0000000000-0.0150.005-0.005-0.035-0.025-0.045-0.035-0.025-0.045-0.025000000000000000000000000000000
00000000000000000000-0.005-0.025-0.0150.0150.0050.0250.0150.0050.0250.00500000000000000000000
00000000000000000000-0.0150.005-0.005-0.035-0.025-0.045-0.035-0.025-0.045-0.02500000000000000000000
10000000001000000000100000000000000000000000000000
01000000000100000000010000000000000000000000000000
00100000000010000000001000000000000000000000000000
00010000000001000000000100000000000000000000000000
00001000000000100000000010000000000000000000000000
00000100000000010000000001000000000000000000000000
00000010000000001000000000100000000000000000000000
00000001000000000100000000010000000000000000000000
00000000100000000010000000001000000000000000000000
00000000010000000001000000000100000000000000000000
00000000000000000000000000000010000000001000000000
00000000000000000000000000000001000000000100000000
00000000000000000000000000000000100000000010000000
00000000000000000000000000000000010000000001000000
00000000000000000000000000000000001000000000100000
00000000000000000000000000000000000100000000010000
00000000000000000000000000000000000010000000001000
00000000000000000000000000000000000001000000000100
00000000000000000000000000000000000000100000000010
00000000000000000000000000000000000000010000000001
11111111110000000000000000000000000000000000000000
00000000001111111111000000000000000000000000000000
00000000000000000000111111111100000000000000000000
00000000000000000000000000000011111111110000000000
00000000000000000000000000000000000000001111111111
10000000001000000000100000000010000000001000000000
01000000000100000000010000000001000000000100000000
00100000000010000000001000000000100000000010000000
00010000000001000000000100000000010000000001000000
00001000000000100000000010000000001000000000100000
00000100000000010000000001000000000100000000010000
00000010000000001000000000100000000010000000001000
00000001000000000100000000010000000001000000000100
00000000100000000010000000001000000000100000000010
00000000010000000001000000000100000000010000000001
];
b=[1.2154-1.21.3154-1.31.3154-1.300000061.6883116968.1818181864.9350649468.1818181871.4285714381.1688311768.1818181884.4155844287.6623376681.1688311781.1688311771.4285714387.6623376668.1818181874.6753246887.6623376668.1818181874.6753246887.6623376681.1688311716016016016016096969696969696969696]';
Aeq=[];
beq=[];
lb=zeros(1,50);
for i=1 :50
ub(i)=inf;
end
[x,z]=linprog(c,A,b,Aeq,beq,lb,ub)