from scipy.optimize import *
import numpy as np
c=[-0.9,-0.45,0.05,-1.4,-0.95,-0.45,-1.9,-1.45,-0.95]
A=[[-0.4,0,0,0.6,0,0,0.6,0,0],[0,-0.7,0,0,0.3,0,0,0.3,0],[-0.2,0,0,-0.2,0,0,0.8,0,0],[0,-0.5,0,0,-0.5,0,0,0.5,0],
[0,0,-0.6,0,0,-0.6,0,0,0.4],[1,1,1,0,0,0,0,0,0],[0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,1,1,1]]
b=[0,0,0,0,0,2000,2500,1200]
bound=((0,2000),(0,2000),(0,2000),(0,2500),(0,2500),(0,2500),(0,1200),(0,1200),(0,1200))
res=linprog(c,A,b,None,None,bound)
a=np.array(np.reshape(res.x,(3,3)))
print("应生产糖果\n甲:%.0f千克"%(a[0,0]+a[1,0]+a[2,0]))
print("乙:%.0f千克"%(a[0,1]+a[1,1]+a[2,1]))
print("丙:%.0f千克"%(a[0,2]+a[1,2]+a[2,2]))
print('3022')
