編程題 #3
來源: POJ (Coursera聲明:在POJ上完成的習題將不會計入Coursera的最后成績。)
注意: 總時間限制: 1000ms 內存限制: 65536kB
描述
寫一個二維數組類 Array2,使得下面程序的輸出結果是:
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;// 在此處補充你的代碼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;}輸入
無
輸出
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,
樣例輸入
1無樣例輸出
12345670,1,2,3,4,5,6,7,8,9,10,11,next0,1,2,3,4,5,6,7,8,9,10,11,代碼:
#include <iostream>#include <cstring>using namespace std;// 在此處補充你的代碼//定義一個類Array2//要重載操作符'[]',支持二維數組下標//要重載操作符 '()'//要進行深度拷貝,建立自己的拷貝構造函數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);};//建立默認構造函數Array2::Array2(){ for (int i = 0; i < 10; i++) for (int j = 0; j < 10; j++) { arr[i][j] = 0; }}//建立構造函數,初始化二維數組arrArray2::Array2(int a, int b){ for (int i = 0; i < a; i++) for (int j = 0; j < b; j++) { arr[i][j] = 0; }}//重載操作符'[]',支持二維數組下標int* Array2::operator [](int i){ return arr[i]; //它是二維數組arr的第i個元素,即第i個小數組的首地址,故返回值類型應該為int* }//重載操作符 '()' int Array2::operator()(int i, int j){ return arr[i][j];}//建立自己的拷貝構造函數,進行深度拷貝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; //說明要重載操作符'[]',支持二維數組下標 for (i = 0; i < 3; ++i) { for (j = 0; j < 4; j++) { cout << a(i, j) << ","; //說明要重載操作符 '()' } cout << endl; } cout << "next" << endl; Array2 b; b = a; //說明要進行深度拷貝 for (i = 0; i < 3; ++i) { for (j = 0; j < 4; j++) { cout << b[i][j] << ","; } cout << endl; } return 0;}