信息安全管理作业 O

#include <iostream>
#include <vector>
#include <cmath>
#include <iomanip>
void output(double ans) {
    int a = static_cast<int>(ans * 10000 + 0.5);
    std::cout << std::fixed << std::setprecision(4) << static_cast<double>(a) / 10000 << std::endl;
}
int main()
{
	std::vector<std::vector<double>> C(3, std::vector<double>(5, 0));
	for (int i = 0; i < 3; i++)
	{
		for (int j = 0; j < 5; j++)
		{
			std::cin >> C[i][j];
		}
	}
	std::vector<std::vector<double>> b(3, std::vector<double>(5, 0));
	for (int i = 0; i < 3; i++)
	{
		for (int j = 0; j < 5; j++)
		{
			std::cin >> b[i][j];
		}
	}
	std::vector<std::vector<double>>w4(125, std::vector<double>(5, 0));
	for (int i = 0; i < 125; i++)
	{
		for (int j = 0; j < 5; j++)
		{
			std::cin >> w4[i][j];
		}
	}
	std::vector<double> w5(5, 0.0);
	for (int i = 0; i < 5; i++)
	{
		std::cin >> w5[i];
	}
	int n;
	std::cin >> n;
	std::vector<std::vector<double>> t1(n, std::vector<double>(3, 0));
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < 3; j++)
		{
			std::cin >> t1[i][j];//第一层输入
		}
	}
	for (int i = 0; i < n; i++)//开始n次循环
	{
		std::vector<std::vector<double>> t2(3, std::vector<double>(5, 0));
		for (int j = 0; j < 3; j++)
		{
			for (int k = 0; k < 5; k++)
			{
				double x = (pow(t1[i][j] - C[j][k], 2)) / (pow(b[j][k], 2));
				t2[j][k]= exp(-x);
			}
			
			
		}
		std::vector<double> t3(125, 1.0);
		/*for (int k = 0; k < 125; k++) {
			for (int j = 0; j < 3; j++) {
				t3[k] *= t2[j][k % 5];//第3层输出
			}
		}*/
		int num = 0;
		for (int j = 0; j < 5; j++)
		{
			for (int k = 0; k < 5; k++)
			{
				for (int e = 0; e < 5; e++)
				{
					t3[num]=t2[0][j] * t2[1][k] * t2[2][e];
					num++;
				}
			}
		}

		std::vector<double> t4(5, 0.0);
		for (int j = 0; j < 5; j++)
		{
			for (int k = 0; k < 125; k++)
			{
				t4[j] += t3[k] * w4[k][j];
			}
		}
		for (int j = 0; j < 5; j++)//归一化
		{
			t4[j] = 1 / (1 + exp(-t4[j]));
		}
		//第5层
		double sum = 0;
		for (int j = 0; j < 5; j++)
		{
			sum += w5 [j] * t4[j];
		}
		//std::cout<< std::fixed << std::setprecision(4)<<1/ (1 + exp(-sum))<<std::endl;
		output(1/ (1 + exp(-sum)));

	}

}