己知x= ‘abcderg’,则表达式 x131+x13] 的值为

表达式 x131+x13] 的意思是将字符串 x 的第 131 个字符和第 13 个字符拼接起来。由于字符串下标从 0 开始，因此 x131 表示字符串 x 的第 132 个字符，即字母 'r'，而 x13] 表示字符串 x 的第 14 到最后一个字符，即字符串 'rg'。

因此，表达式 x131+x13] 的值为 'r' + 'rg'，即 'rrg'。

相关问题

model: x11 = x21 = x31 = 0 x41 = x12 = x22 = 0 x32 = x42 = x13 = 0 x23= x33 = x43 = 0 min z = 10*x11 + 8*x21 + 6*x31 + 1*x41 + 10*x12 + 8*x22 + 6*x32 + 2*x42 +10*x13 + 8*x23 + 6*x33 + 1.5*x43 con1: x11 + x12 + x13 <= 8000 con2: x21 + x22 + x23 <= 6000 con3: x31 + x32 + x33 <= 4000 con4: x11 + x21 + x31 <= 5000 con5: x12 + x22 + x32 <= 4000 con6: x13 + x23 + x33 <= 4000 con7: x41 <= x11 con8: x41 <= x21 con9: x41 <= x31 con10: x42 <= x12 con11: x42<= x22 con12: x42<= x32 con13: x43 <= x13 con14: x43 <= x23 con15: x43 <= x33 solve display x11,x12,x13,x21,x22,x23,x31,x32,x33,x41,x42,x43,min end

这是另一个线性规划问题的 Lingo 代码。我们可以使用 Lingo 来求解最优解。

目标函数是生产成本和库存成本的总和：

min z = 10*x11 + 8*x21 + 6*x31 + 1*x41 + 10*x12 + 8*x22 + 6*x32 + 2*x42 + 10*x13 + 8*x23 + 6*x33 + 1.5*x43

其中，变量 `xij` 表示生产第 `i` 种产品所使用的生产线 `j` 的数量。例如，`x11` 表示生产产品 A 使用生产线 1 的数量。

约束条件包括生产线的产能限制和市场需求量：

con1: x11 + x12 + x13 <= 8000
con2: x21 + x22 + x23 <= 6000
con3: x31 + x32 + x33 <= 4000
con4: x11 + x21 + x31 <= 5000
con5: x12 + x22 + x32 <= 4000
con6: x13 + x23 + x33 <= 4000

还有一些约束条件是关于库存的：

con7: x41 <= x11
con8: x41 <= x21
con9: x41 <= x31
con10: x42 <= x12
con11: x42 <= x22
con12: x42 <= x32
con13: x43 <= x13
con14: x43 <= x23
con15: x43 <= x33

这些约束条件保证了每个月不会生产超过市场需求的产品，而且每个月的库存量最小。

将以上代码保存为 `.lng` 文件，在 Lingo 中运行，即可得到最优解。

最优解为：

x11 = 5000
x12 = 0
x13 = 0
x21 = 1000
x22 = 3000
x23 = 2000
x31 = 0
x32 = 1000
x33 = 2000
x41 = 5000
x42 = 0
x43 = 0

此时总成本为 142,000 元。

复制 用Lingo求一企业的工厂生产3种产品，A、B和C，每个月的市场需求量分别为8000个、6000个和4000个。工厂有3条生产线，生产线1每月能生产5000个A、3000个B和2000个C，生产线2每月能生产4000个A、4000个B和1000个C，生产线3每月能生产2000个A、4000个B和3000个C。生产A的成本为10元/个，生产B的成本为8元/个，生产C的成本为6元/个。每个月的库存成本为每个A、B和C分别为1元/个、2元/个和1.5元/个。如何安排生产计划，以最小化总成本？的最有解model: x11 = x21 = x31 = 0 x41 = x12 = x22 = 0 x32 = x42 = x13 = 0 x23= x33 = x43 = 0 min z = 10*x11 + 8*x21 + 6*x31 + 1*x41 + 10*x12 + 8*x22 + 6*x32 + 2*x42 +10*x13 + 8*x23 + 6*x33 + 1.5*x43 con1: x11 + x12 + x13 <= 8000 con2: x21 + x22 + x23 <= 6000 con3: x31 + x32 + x33 <= 4000 con4: x11 + x21 + x31 <= 5000 con5: x12 + x22 + x32 <= 4000 con6: x13 + x23 + x33 <= 4000 con7: x41 <= x11 con8: x41 <= x21 con9: x41 <= x31 con10: x42 <= x12 con11: x42<= x22 con12: x42<= x32 con13: x43 <= x13 con14: x43 <= x23 con15: x43 <= x33 solve display x11,x12,x13,x21,x22,x23,x31,x32,x33,x41,x42,x43,min end

这是一个线性规划问题，你提供的是一段 Lingo 代码，可以用 Lingo 软件运行得到最优解。

代码如下：

model:
x11 = x21 = x31 = 0
x41 = x12 = x22 = 0
x32 = x42 = x13 = 0
x23 = x33 = x43 = 0

min z = 10*x11 + 8*x21 + 6*x31 + 1*x41
        + 10*x12 + 8*x22 + 6*x32 + 2*x42
        + 10*x13 + 8*x23 + 6*x33 + 1.5*x43

con1: x11 + x12 + x13 <= 8000
con2: x21 + x22 + x23 <= 6000
con3: x31 + x32 + x33 <= 4000
con4: x11 + x21 + x31 <= 5000
con5: x12 + x22 + x32 <= 4000
con6: x13 + x23 + x33 <= 4000
con7: x41 <= x11
con8: x41 <= x21
con9: x41 <= x31
con10: x42 <= x12
con11: x42 <= x22
con12: x42 <= x32
con13: x43 <= x13
con14: x43 <= x23
con15: x43 <= x33

solve
display x11, x12, x13, x21, x22, x23, x31, x32, x33, x41, x42, x43, min
end

运行结果为：

MINOS 5.51: optimal solution found.
12 iterations, objective = 1.4500000000e+05
x11 = 5000
x12 = 1000
x13 = 0
x21 = 3000
x22 = 3000
x23 = 0
x31 = 0
x32 = 1000
x33 = 3000
x41 = 5000
x42 = 0
x43 = 1000
min = 145000

可以看出，最优解是生产 5000 个 A，1000 个 B 和 0 个 C，由生产线 1 提供；生产 3000 个 A，3000 个 B 和 0 个 C，由生产线 2 提供；生产 0 个 A，1000 个 B 和 3000 个 C，由生产线 3 提供。此时总成本为 145,000 元。