please use matlab. def ispythag(a, b, c): if a**2 + b**2 == c**2: return 1 else: return 0
时间: 2024-01-17 14:55:44 浏览: 33
Here is an implementation of the function in MATLAB:
function result = ispythag(a, b, c)
if a^2 + b^2 == c^2
result = 1;
else
result = 0;
end
end
You can call this function with three input arguments a, b, and c to determine if they form a Pythagorean triple. The function returns 1 if the inputs satisfy the Pythagorean theorem, and 0 otherwise. Here's an example usage:
>> ispythag(3, 4, 5)
ans =
1
This indicates that (3, 4, 5) is a Pythagorean triple, since 3^2 + 4^2 = 5^2.
相关问题
c语言1*2*3*...(2n+1)用for循环
以下是使用for循环计算1*3*5*...*(2n+1)的C语言代码:
```c
#include <stdio.h>
int main() {
int n;
int product = 1;
printf("1*3*5*...*(2n+1)\n");
printf("Please input 'n' to calculate:");
scanf("%d", &n);
for (int i = 0; i <= n; i++) {
product *= (2 * i + 1);
}
printf("PRODUCT = %d\n", product);
return 0;
}
```
代码解释:
1. 首先,我们定义了变量n和product,其中n表示要计算的项数,product表示最终的积。
2. 然后,我们使用printf函数输出提示信息,要求用户输入n。
3. 接着,我们使用scanf函数读取用户输入的n。
4. 然后,我们使用for循环计算积。在每次循环中,我们将2i+1乘到product中。
5. 最后,我们使用printf函数输出计算结果。
计算1*3*5*7*....*(2n+1)
以下是计算1*3*5*7*....*(2n+1)的C语言代码:
```c
#include <stdio.h>
int n;
int product = 1;
int main() {
int product_temporary;
int loop_counter;
printf("1*3*5*...*(2n+1)\n");
printf("Please input 'n' to calculate:");
scanf("%d",&n);
for (int loop_counter = 0; loop_counter < n; ++loop_counter) {
product_temporary = 2 * loop_counter + 1;
product = product * product_temporary;
}
printf("PRODUCT = %d \n",product);
return 0;
}
```
运行结果如下:
```
1*3*5*...*(2n+1)
Please input 'n' to calculate:5
PRODUCT = 945
```
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