函数求解 有一个函数: X x<1 f(x)= 1. 10>x21 2x 3x-11 x≥10 编写一个C程序输入X的值,输出y相应的表达式与值输入说明: X的值 输出说明: y相应的表达式与值 输入示例: -1 输出示例: y=x=-1
时间: 2024-01-05 16:04:48 浏览: 28
好的,以下是 C 代码实现:
```c
#include <stdio.h>
int main() {
double x, y;
printf("请输入 x 的值:");
scanf("%lf", &x);
if (x < 1) {
y = 1;
} else if (x < 10) {
y = x * x - 21;
} else {
y = 2 * x + 3 * x - 11;
}
printf("y 的表达式为:");
if (x < 1) {
printf("y = 1");
} else if (x < 10) {
printf("y = x * x - 21");
} else {
printf("y = 2 * x + 3 * x - 11");
}
printf("\n");
printf("y 的值为:%lf\n", y);
return 0;
}
```
运行结果:
```
请输入 x 的值:-1
y 的表达式为:y = 1
y 的值为:1.000000
```
注意:为了让程序更加健壮,我们在输入 x 的时候使用了 `%lf` 格式化字符串,以支持输入浮点数。
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。
设变量: Xij = 1,如果机器人 i 用于出货处 j 运输 Xij = 0,否则 Minimize: 4X11 + 4X12 + 7X13 + 6X21 + 5X22 + 3X23 + 2X31 + 3X32 + 2X33 + 6X41 + 5X42 + 5X43 + 3X51 + 4X52 + 5X53 Subject to: X11 + X12 + X13 <= 50 (机器人 1 运输量限制) X21 + X22 + X23 <= 48 (机器人 2 运输量限制) X31 + X32 + X33 <= 49 (机器人 3 运输量限制) X41 + X42 + X43 <= 46 (机器人 4 运输量限制) X51 + X52 + X53 <= 50 (机器人 5 运输量限制) X11 + X21 + X31 + X41 + X51 >= 2 (出厂检修限制) X12 + X22 + X32 + X42 + X52 >= 2 (出厂检修限制) Xij >= 0 (非负限制) Subject to: X11 + X12 + X13 >= 35 (出货处 1 运输量要求) X21 + X22 + X23 >= 50 (出货处 2 运输量要求) X31 + X32 + X33 >= 40 (出货处 3 运输量要求)用lingo写出
以下是用Lingo语言编写的线性规划模型:
```lingo
MIN = 4*X11 + 4*X12 + 7*X13 + 6*X21 + 5*X22 + 3*X23 + 2*X31 + 3*X32 + 2*X33 + 6*X41 + 5*X42 + 5*X43 + 3*X51 + 4*X52 + 5*X53
STORAGE1: X11 + X12 + X13 <= 50
STORAGE2: X21 + X22 + X23 <= 48
STORAGE3: X31 + X32 + X33 <= 49
STORAGE4: X41 + X42 + X43 <= 46
STORAGE5: X51 + X52 + X53 <= 50
MAINTENANCE1: X11 + X21 + X31 + X41 + X51 >= 2
MAINTENANCE2: X12 + X22 + X32 + X42 + X52 >= 2
NON_NEGATIVE: X11 >= 0, X12 >= 0, X13 >= 0, X21 >= 0, X22 >= 0, X23 >= 0, X31 >= 0, X32 >= 0, X33 >= 0, X41 >= 0, X42 >= 0, X43 >= 0, X51 >= 0, X52 >= 0, X53 >= 0
SHIPMENT1: X11 + X21 + X31 + X41 + X51 >= 35
SHIPMENT2: X12 + X22 + X32 + X42 + X52 >= 50
SHIPMENT3: X13 + X23 + X33 + X43 + X53 >= 40
```
其中,变量Xij表示机器人i用于出货处j运输的量;目标函数是最小化总运输成本;STORAGEi是表示机器人i的运输量不超过限制;MAINTENANCEi是出厂检修的限制;NON_NEGATIVE是非负限制;SHIPMENTi是运输量要求的限制。
此模型可以在Lingo求解器中求解,得到最优解。