Aug 16

C++矩阵类(class Matrix) 不指定

felix021 @ 2008-8-16 00:41 [IT » 程序设计] 评论(3) , 引用(0) , 阅读(12169) | Via 本站原创 | |
@ 2009-05-25 其实早就发现了这个代码有严重的内存泄露问题,不过懒得改了。。。

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实现重载了等于=  加法+  减法-  矩阵乘法*  矩阵幂^(使用快速幂算法)  输出<<  输入>>,用起来会比较方便了,不过就是有点长。

查了好久快速幂算法,网上居然没有现成的,于是只好自己再推一遍,先贴出是整数的快速幂算法:
int Pow(int x, int y){
    int ans, b, i;
    if(y == 0)return 1;
    if(y < 0) return -1;
    for(i = y, b = 0; i > 0; i >>= 1) b++;
    for(ans = 1, i = b - 1; i >= 0; i--){
        if(y & (1 << i)){
            ans = ans * ans * x;
        }else{
            ans = ans * ans;
        }
    }
    return ans;
}

下面是矩阵类的代码:
#include<iostream>
#include<math.h>
using namespace std;

template <typename T>
class Matrix{
//class Matrix by felix021
//usage: Matrix<typename> varname(row, col[, mod]);
//如果typename不是short/int/long/long long
//那么取模运算需要包含math.h头文件(使用fmod函数)
//取模运算的代码仅在 + - * 三个函数内.

    public:
        T **data;
        Matrix *temp;
        int row, col, modnum;
        //无参构造函数,没有给定矩阵的大小时,输出错误提示
        Matrix(){
            printf("Size parameter invalid!\n");
        }
        //构造函数, r是行数,c是列数,若给定m>0, 则在求和/差/积时对m取模
        Matrix(int r, int c, int m = 0){
            int i;
            temp = NULL;
            row = r, col = c;
            data = new T *[r]; //分配一个 T *[r]类型的数组给data
            for(i = 0; i < r; i++){
                data[i] = new T [c];  //给data下的每个指针分配内存, 类型为 T [c]
            }
            if(m > 0) modnum = m;
            else modnum = 0;
        }
        //析构函数,用于释放分配的内存
        ~Matrix(){
            int i;
            if(temp != NULL){delete temp; }
            for(i = 0; i < row; i++) delete data[i];
            delete data;
        }
        //赋值,给第i行第j列的元素赋值为 a
        void assign(int i, int j, const T a){
            if(i < 0 || i >= row || j < 0 || j >= col) {
                printf("Invalid row/column number.\n");
                return;
            }else{
                data[i][j] = a;
            }
        }
        //取出元素,给第i行第j列的元素的值
        T at(int i, int j){
            if(i < 0 || i >= row || j < 0 || j >= col) {
                printf("Invalid row/column number.\n");
                return;
            }else{
                return data[i][j];
            }
        }
        //重载<<运算符,可用cout输出
        friend inline ostream & operator<<(ostream &os, const Matrix &a){
            int i, j;
            for(i = 0; i < a.row; i++){
                for(j = 0; j < a.col - 1; j++){
                    os << a.data[i][j] << " ";
                }
                os << a.data[i][j];
                os << endl;
            }
            return os;
        }
        //重载>>运算符,可用cin输入
        friend inline istream & operator>>(istream &is, const Matrix &a){
            int i, j;
            for(i = 0; i < a.row; i++){
                for(j = 0; j < a.col; j++){
                    is >> a.data[i][j];
                }
            }
            return is;
        }
        //重载=运算符,要求两个矩阵的大小相同
        Matrix & operator = (const Matrix &a){
            int i, j;
            if(row != a.row || col != a.col){
                printf("Unmatch Matrix!\n");
                return *this;
            }
            for(i = 0; i < row; i++){
                for(j = 0; j < col; j++){
                    data[i][j] = a.data[i][j];
                    if(modnum > 0){
                        data[i][j] = mod(data[i][j]);
                    }
                }
            }
            return *this;
        }
        //重载+运算符,要求两个矩阵的大小相同
        Matrix & operator + (const Matrix &a){
            int i, j;
            if(row != a.row || col != a.col){
                printf("Unmatch Matrix!\n");
                return *this;
            }
            if(temp != NULL) delete temp;
            temp = new Matrix(row, col, modnum);
            for(i = 0; i < row; i++){
                for(j = 0; j < col; j++){
                    temp->data[i][j] = data[i][j] + a.data[i][j];
                    if(modnum > 0){
                        temp->data[i][j] = mod(temp->data[i][j]);
                    }
                }
            }
            return *temp;
        }
        //重载-运算符,要求两个矩阵的大小相同
        Matrix & operator - (const Matrix &a){
            int i, j;
            if(row != a.row || col != a.col){
                printf("Unmatch Matrix!\n");
                return *this;
            }
            if(temp != NULL) delete temp;
            temp = new Matrix(row, col, modnum);
            for(i = 0; i < row; i++){
                for(j = 0; j < col; j++){
                    temp->data[i][j] = data[i][j] - a.data[i][j];
                    if(modnum > 0){
                        temp->data[i][j] = mod(temp->data[i][j] + modnum);
                    }
                }
            }
            return *temp;
        }
        //重载*运算符,要求矩阵a的列数等于b的行数
        Matrix & operator * (const Matrix &a){
            int i, j, k;
            T tmp;
            if(col != a.row){
                printf("Unmatch Matrix!\n");
                return *this;
            }
            if(temp != NULL) delete temp;
            temp  = new Matrix(row, a.col, modnum);
            for(i = 0; i < row; i++){
                for(j = 0; j < a.col; j++){
                    tmp = 0;
                    for(k = 0; k < a.row; k++){
                        tmp += data[i][k] * a.data[k][j];
                        if(modnum > 0){
                            tmp = mod(tmp);
                        }
                    }
                    temp->data[i][j] = tmp;
                }
            }
            return *temp;
        }
        //重载^运算符,要求矩阵的列数等于行数
        Matrix & operator ^ (const int a){
            int i, j;
            if(row != col){
                printf("No n*n matrix!\n");
                return *this;
            }
            if(a < 0){
                printf("Invalid a(%d)!\n", a);
                return *this;
            }
            if(temp != NULL) delete temp;
            temp  = new Matrix(row, col, modnum);
            for(i = 0; i < row; i++){ //单位方阵
                for(j = 0; j < col; j++){
                    if(i == j )temp->data[i][j] = 1;
                    else temp->data[i][j] = 0;
                }
            }
            for(i = a, j = 0; i > 0; i >>= 1) j++;
            for(i = j - 1; i >= 0; i--){
                if(a & (1 << i)){
                    *temp = *temp * *temp * (*this);
                }else{
                    *temp = *temp * *temp;
                }
            }
            return *temp;
        }
        //重载求模函数,对int, long, long long, float, double都有效
        int mod(int i){
            return i % modnum;
        }
        long mod(long i){
            return i % modnum;
        }
        long long mod(long long i){
            return i % modnum;
        }
        float mod(float i){
            return fmod(i, modnum);
        }
        double mod(double i){
            return fmod(i, modnum);
        }
};

int main(){
    int i, j;
    Matrix <int> t(2, 2), t1(2, 2), t2(2, 2), a(5, 10, 7), b(10, 5), c(5, 5);
    cout << "(cin/cout test)\nPlease input 2 (2*2) Matrix: \n";
    cin >> t;
    cout << t << endl;
    t1 = t * t * t * t * t;
    t2 = t ^ 5;
    cout << "t^5 = \n" << t1 << endl << t2 << endl;
    getchar();
    
    for(i = 0; i < 5; i++){
        for(j = 0; j < 10; j++){
            a.assign(i, j, 1);
            b.assign(j, i, 2);
        }
    }
    c = a * b;
    cout << "Matrix a:" << endl << a << endl;
    cout << "Matrix b:" << endl << b << endl;
    cout << "Matrix a * b:" << endl << c << endl;
    getchar();
    return 0;
}




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转载请注明出自 ,如是转载文则注明原出处,谢谢:)
RSS订阅地址: https://www.felix021.com/blog/feed.php
coco
2009-11-1 20:05
很强,谢谢
dxhs
2009-1-18 20:49
pig
呵呵
2008-11-4 22:28
觉得有用  但是不是很有用!还是很感谢!
felix021 回复于 2008-11-5 12:28
代码貌似有内存泄露的可能性。。。囧。。
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