在线笔试模板 c++

1.矩阵运算

#include <iostream>
#include <algorithm>

#include <vector>
using namespace std;

#define Debug_ 1

//galobal value
#define N  2 //matrix inv

vector<vector<int>> Matrixmulti(vector<vector<int>> a, vector<vector<int>> b)
{
	int RowA = a.size();
	int ColA = a[0].size();
	int RowB = b.size();
	int ColB = b[0].size();

	vector<vector<int>> c(RowA, vector<int>(ColB, 0));
	int s=0;

#if Debug_
	if (ColA != RowB)
	{
		cout << "Matrix is not correct!" << endl;
		return c;
	}
#endif
	for (int i = 0; i < RowA; i++)
	{
		for (int m = 0; m <ColB; m++)
		{
			for (int j = 0; j < ColA; j++)
			{
				s = s + a[i][j] * b[j][m];
			}
			c[i][m] = s;
			s = 0;
		}
	}
	
	return c;
}

vector<vector<int>> matrix_transpose(vector<vector<int>> a)
{
	int m = a.size();
	int n = a[0].size();
	vector<vector<int>> c= a;

#if Debug_
	cout << "矩阵行列为" << m<<" " << n << endl;
	if (m != n)
	{
		cout << "矩阵行列不一致！" << endl;
		return c;
	}
#endif

	int i, j;
	for (i = 1; i<m; i++)
	{
		for (j = 0; j<i; j++)
			std::swap(c[i][j], c[j][i]);
	}
	return c;
}


//LUP分解
void LUP_Descomposition(double A[N*N], double L[N*N], double U[N*N], int P[N])
 {
     int row = 0;
     for (int i = 0; i<N; i++)
     {
         P[i] = i;
     }
     for (int i = 0; i<N - 1; i++)
     {
         double p = 0.0;
         for (int j = i; j<N; j++)
         {
             if (abs(A[j*N + i])>p)
             {
                 p = abs(A[j*N + i]);
                 row = j;
             }
         }
         if (0 == p)
         {
             cout << "矩阵奇异，无法计算逆" << endl;
             return;
         }

         //交换P[i]和P[row]
         int tmp = P[i];
         P[i] = P[row];
         P[row] = tmp;

         double tmp2 = 0.0;
        for (int j = 0; j<N; j++)
         {
             //交换A[i][j]和 A[row][j]
             tmp2 = A[i*N + j];
            A[i*N + j] = A[row*N + j];
             A[row*N + j] = tmp2;
         }

         //以下同LU分解
         double u = A[i*N + i], l = 0.0;
         for (int j = i + 1; j<N; j++)
         {
             l = A[j*N + i] / u;
             A[j*N + i] = l;
             for (int k = i + 1; k<N; k++)
             {
                 A[j*N + k] = A[j*N + k] - A[i*N + k] * l;
             }
         }
     }

     //构造L和U
     for (int i = 0; i<N; i++)
     {
         for (int j = 0; j <= i; j++)
         {
             if (i != j)
             {
                 L[i*N + j] = A[i*N + j];
             }
             else
             {
                 L[i*N + j] = 1;
             }
         }
         for (int k = i; k<N; k++)
         {
             U[i*N + k] = A[i*N + k];
         }
     }

 }
double * LUP_Solve(double L[N*N], double U[N*N], int P[N], double b[N])
 {
     double *x = new double[N]();
     double *y = new double[N]();

     //正向替换
     for (int i = 0; i < N; i++)
    {
         y[i] = b[P[i]];
         for (int j = 0; j < i; j++)
         {
             y[i] = y[i] - L[i*N + j] * y[j];
         }
     }
     //反向替换
     for (int i = N - 1; i >= 0; i--)
     {
         x[i] = y[i];
         for (int j = N - 1; j > i; j--)
         {
             x[i] = x[i] - U[i*N + j] * x[j];
         }
         x[i] /= U[i*N + i];
     }
     return x;
 }
/* 转置，即循环处理所有环 */
/* 后继 */
int getNext(int i, int m, int n)
{
  return (i%n)*m + i / n;
 }

 /* 前驱 */
 int getPre(int i, int m, int n)
 {
   return (i%m)*n + i / m;
 }

 /* 处理以下标i为起点的环 */
 void movedata(double *mtx, int i, int m, int n)
 {
	   double temp = mtx[i]; // 暂存
	   int cur = i;    // 当前下标
	   int pre = getPre(cur, m, n);
	   while (pre != i)
		   {
		     mtx[cur] = mtx[pre];
		     cur = pre;
		     pre = getPre(cur, m, n);
		   }
	   mtx[cur] = temp;
 }
 void transpose(double *mtx, int m, int n)
{
	   for (int i = 0; i<m*n; ++i)
	   {
	    int next = getNext(i, m, n);
	     while (next > i) // 若存在后继小于i说明重复,就不进行下去了（只有不重复时进入while循环）
	       next = getNext(next, m, n);
	     if (next == i)  // 处理当前环
	       movedata(mtx, i, m, n);
	   }
 }

vector<vector<double>> LUP_solve_inverse(vector<vector<int>> a)  //LUP 方程
 {
	     //创建矩阵A的副本，注意不能直接用A计算，因为LUP分解算法已将其改变

	int m = a.size();
	int n = a[0].size();
	vector<vector<double>> out_ (m,vector<double>(n,0));
	
	
	double *A_mirror = new double[N*N]();
	double *inv_A = new double[N*N]();//最终的逆矩阵（还需要转置）
	double *inv_A_each = new double[N]();//矩阵逆的各列
	//double *B    =new double[N*N]();
		double *b = new double[N]();//b阵为B阵的列矩阵分量
	
		for (int i = 0; i<N; i++)
		{
			double *L = new double[N*N]();
			double *U = new double[N*N]();
			int *P = new int[N]();
		
			//构造单位阵的每一列
			for (int i = 0; i<N; i++)
			{
				b[i] = 0;
			}
			b[i] = 1;
		
			//每次都需要重新将A复制一份
			for (int i = 0; i<N*N; i++)
			{
				A_mirror[i] = (double) a[i/N][i%N];
		    }
		
			LUP_Descomposition(A_mirror, L, U, P);
		
			inv_A_each = LUP_Solve(L, U, P, b);
			memcpy(inv_A + i*N, inv_A_each, N * sizeof(double));//将各列拼接起来
		}

	transpose(inv_A, N, N);//由于现在根据每列b算出的x按行存储，因此需转置
	
	for (int i = 0; i<N*N; i++)
	{
		out_[i / N][i%N] = inv_A[i];
	}

	return out_;
 }


int main(void)
{
	int NumOfRowA=2, NumOfColA=2, NumOfRowB=2, NumOfColB=2;
	cin>> NumOfRowA >> NumOfColA>> NumOfRowB >> NumOfColB;

	vector<vector<int>> a(NumOfRowA, vector<int>(NumOfColA, 0));
	vector<vector<int>> b(NumOfRowB, vector<int>(NumOfColB, 0));
	//output
	vector<vector<int>> a_mul_b(NumOfRowA, vector<int>(NumOfColB, 0));
	vector<vector<int>> a_trans(NumOfRowA, vector<int>(NumOfRowA, 0));

	vector<vector<double>> a_inv(NumOfRowA, vector<double>(NumOfRowA, 0));

	cout << "请输入A矩阵：" << endl;
	//input

	for (int i = 0; i < NumOfRowA; i++)
	{
		for (int j = 0; j < NumOfColA; j++)
		{
			cin >> a[i][j];
			cout << a[i][j] << " ";
		}
		cout <<endl;
	}

	cout << "请输入B矩阵" << endl;

	for (int j = 0; j < NumOfColB; j++)
	{
		for (int m = 0; m < NumOfColB; m++)
		{
			cin >> b[j][m];
			cout << b[j][m] << " ";
		}
		cout << endl;
	}
	
	a_mul_b = Matrixmulti(a, b);

	cout << "矩阵乘法的结果：" << endl;

	for (int i = 0; i < NumOfRowA; i++)
	{
		for (int j = 0; j < NumOfColB; j++)
		{
			cout << a_mul_b[i][j] << "\t";
		}
		cout << endl;
	}

	a_trans = matrix_transpose(a);

	cout << "矩阵A转置的结果：" << endl;

	for (int i = 0; i < NumOfRowA; i++)
	{
		for (int j = 0; j < NumOfColA; j++)
		{
			cout << a_trans[i][j] << "\t";
		}
		cout << endl;
	}

	a_inv = LUP_solve_inverse(a);

	cout << "矩阵A求逆的结果：" << endl;

	for (int i = 0; i < NumOfRowA; i++)
	{
		for (int j = 0; j < NumOfRowA; j++)
		{
			cout << a_inv[i][j] << "\t";
		}
		cout << endl;
	}
	

	while (1);

	return 0;
}

2.一步状态转移矩阵

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

#define udirection 1 //无向图
#define direction  0 //有向图
#define Freach     0 //第一次到达
#define FindAll    1 //遍历所有节点

vector<vector<int>> Matrixmulti(vector<vector<int>> a, vector<vector<int>> b)
{
	int RowA = a.size();
	int ColA = a[0].size();
	int RowB = b.size();
	int ColB = b[0].size();

	vector<vector<int>> c(RowA, vector<int>(ColB, 0));
	int s=0;

	if (ColA != RowB)
	{
		cout << "Matrix is not correct!" << endl;
		return c;
	}
	for (int i = 0; i < RowA; i++)
	{
		for (int m = 0; m <ColB; m++)
		{
			for (int j = 0; j < ColA; j++)
			{
				s = s + a[i][j] * b[j][m];
			}
			c[i][m] = s;
			s = 0;
		}
	}
	
	return c;
}

int main(void)
{
	int NumOfRowA = 4, NumOfColA = 4;

	cin >> NumOfRowA;
		
	vector<vector<int>> a(NumOfRowA, vector<int>(NumOfRowA, 0));// row(start) col(end)
	vector<vector<int>> a_mul_b(NumOfRowA, vector<int>(NumOfRowA, 0));
	vector<int> findip(NumOfRowA);


	cout << "连接关系：" << endl;  //4节点 3条线
	for (int i = 0; i < NumOfRowA - 1; i++)
	{
		int one, two;
		cin >> one >> two; //两个节点的连接线
#if udirection
		a[one-1][two-1] = 1;
		a[two-1][one-1] = 1;
#endif
#if direction
		a[one - 1][two - 1] = 1;
#endif

	}
#if FindAll
	/*for (int i = 0; i < NumOfRowA - 1; i++)
		a[i][i]=1;*/
#endif
	//input
	cout << "一步状态转移矩阵为:" << endl;

	for (int i = 0; i < NumOfRowA; i++)
	{
		for (int j = 0; j < NumOfRowA; j++)
		{
			cout << a[i][j] << " ";
		}
		cout <<endl;
	}
	cout << endl;

	a_mul_b = a;
	for (int step =1;step<5; step++)
	{
		cout << "step "<<step<<":"<< endl<<endl;

		for (int i = 0; i < NumOfRowA; i++)
		{
			for (int j = 0; j < NumOfRowA; j++)
			{
				cout << a[i][j] << "\t";
			}
			cout << endl;
		}
		
#if Freach
		if (a[0][3] == 1)
		{
			cout << "从节点 1 出发第一次到达节点 4 经过 " << step << " 步;" << endl;
			break;
		}
#endif

#if FindAll
		int count_ = 0;
		//for (int i = 0; i < NumOfRowA; i++)
		//	if (a[0][i] != 0)  //a[0][i]表示从节点1出发
		//		findip[i]++;
		//
		for (int i = 0; i < NumOfRowA; i++)
			if (a[0][i] != 0)  //a[0][i]表示从节点1出发
				count_++;
		
		if (count_ == NumOfRowA)
		{
			cout << "从节点 1 出发遍历全部节点最小需要 " << step << " 步;" << endl;
		//	break;
		}
#endif

			a = Matrixmulti(a_mul_b,a);	

	}
	


	while (1);

	return 0;
}

字符串子串

#include<iostream>
#include <string>
#include<math.h>
#include<vector>
using namespace std;

int main()
{
	int k;
	string a, b;

	cin >> k;
	cin >> a;
	cin >> b;
	int len = a.length();

	vector<string> str;

	for (int i = 0; i < len - k+1; i++)
	{
		string s = a.substr(i, k);
	//	cout << s << endl;
		bool fh = false;
		for (int j = 0; j < str.size(); j++)
			if (s == str[j])
				fh = true;
		if (!fh)
			str.push_back(s);
	}

	for (int j = 0; j < str.size(); j++)
		cout << str[j] << endl;

	int num = 0;
	for (int j = 0; j < str.size(); j++)
	{
		for (int i = 0; i < b.length() - k+1; i++)
		{
			string s = b.substr(i, k);
			if (s == str[j])
				num++;
		}
		
	}

	cout << num;

	//while (1);

	return 0;

}