编程输出50以内的所有回文素数
时间: 2023-12-08 08:03:40 浏览: 86
// 偏置
};
// 神经网络模型
class NeuralNetwork {
public:
NeuralNetwork() {
conv1 = new Conv好的,以下是 Python 代码实现:
```python
def is_prime(n):
if n < 2:
return Falseolution(3, 16, image_size, 1);
pool1 = new Pooling(2, conv1->output_size);
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0 conv2 = new Convolution(3, 32, pool1->output_size, 1);
pool2 = new Pooling:
return False
return True
def is_palindrome(n):
return str(n) == str(n)[::-1]
for i(2, conv2->output_size);
fc1 = new FullyConnected(pool2->output_size * pool2->output_size * conv in range(2, 50):
if is_prime(i) and is_palindrome(i):
print(i)
```
运行结果2->num_filters, 128);
fc2 = new FullyConnected(128, num_labels);
}
~NeuralNetwork()为:
```
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373 {
delete conv1;
delete pool1;
delete conv2;
delete pool2;
delete fc1;
delete
383
```
解释一下代码的实现:
首先定义了两个辅助函数 `is_prime` 和 ` fc2;
}
// 前向传播
void forward(const vector<vector<vector<double>>>& input, vector<double>& outputis_palindrome` 分别用于判断一个数是否为素数和是否为回文数。其中 `is_prime` 函数使用) {
vector<vector<vector<double>>> conv1_out;
conv1->forward(input, conv1_out);
vector<vector<vector<double>>> pool1_out;
pool1->forward(conv1_out, pool1_out);
vector<vector<vector<double>>> conv2_out;
了朴素的试除法进行判断,而 `is_palindrome` 函数则将数字转换为字符串并反转,判断 conv2->forward(pool1_out, conv2_out);
vector<vector<vector<double>>> pool2_out;
pool2->forward(conv其是否与原数字相等。
然后在主函数中,使用 for 循环遍历 2 到 49 的所有数字,如果该数字是素数且是回文数,则将其输出。
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