编程题 #3
来源: POJ (Coursera声明:在POJ上完成的习题将不会计入Coursera的最后成绩。)
注意: 总时间限制: 1000ms 内存限制: 65536kB
描述
写一个二维数组类 Array2,使得下面程序的输出结果是:
0,1,2,3,
4,5,6,7,
8,9,10,11,
next
0,1,2,3,
4,5,6,7,
8,9,10,11,
程序:
1234567891011121314151617181920212223242526#include <iostream>#include <cstring>using namespace std;// 在此处补充你的代码int main() { Array2 a(3,4); int i,j; for( i = 0;i < 3; ++i ) for( j = 0; j < 4; j ++ ) a[i][j] = i * 4 + j; for( i = 0;i < 3; ++i ) { for( j = 0; j < 4; j ++ ) { cout << a(i,j) << ","; } cout << endl; } cout << "next" << endl; Array2 b; b = a; for( i = 0;i < 3; ++i ) { for( j = 0; j < 4; j ++ ) { cout << b[i][j] << ","; } cout << endl; } return 0;}输入
无
输出
0,1,2,3,
4,5,6,7,
8,9,10,11,
next
0,1,2,3,
4,5,6,7,
8,9,10,11,
样例输入
1无样例输出
12345670,1,2,3,4,5,6,7,8,9,10,11,next0,1,2,3,4,5,6,7,8,9,10,11,代码:
#include <iostream>#include <cstring>using namespace std;// 在此处补充你的代码//定义一个类Array2//要重载操作符'[]',支持二维数组下标//要重载操作符 '()'//要进行深度拷贝,建立自己的拷贝构造函数class Array2{PRivate: int arr[10][10];public: Array2(); Array2(int, int); int* Operator [](int i); int operator()(int i, int j); Array2(Array2& c);};//建立默认构造函数Array2::Array2(){ for (int i = 0; i < 10; i++) for (int j = 0; j < 10; j++) { arr[i][j] = 0; }}//建立构造函数,初始化二维数组arrArray2::Array2(int a, int b){ for (int i = 0; i < a; i++) for (int j = 0; j < b; j++) { arr[i][j] = 0; }}//重载操作符'[]',支持二维数组下标int* Array2::operator [](int i){ return arr[i]; //它是二维数组arr的第i个元素,即第i个小数组的首地址,故返回值类型应该为int* }//重载操作符 '()' int Array2::operator()(int i, int j){ return arr[i][j];}//建立自己的拷贝构造函数,进行深度拷贝Array2::Array2(Array2& c){ for (int i = 0; i < 10; i++) for (int j = 0; j < 10; j++) { arr[i][j] = c.arr[i][j]; }}int main() { Array2 a(3, 4); int i, j; for (i = 0; i < 3; ++i) for (j = 0; j < 4; j++) a[i][j] = i * 4 + j; //说明要重载操作符'[]',支持二维数组下标 for (i = 0; i < 3; ++i) { for (j = 0; j < 4; j++) { cout << a(i, j) << ","; //说明要重载操作符 '()' } cout << endl; } cout << "next" << endl; Array2 b; b = a; //说明要进行深度拷贝 for (i = 0; i < 3; ++i) { for (j = 0; j < 4; j++) { cout << b[i][j] << ","; } cout << endl; } return 0;}