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- [12pts] Suppose we want to read an array size from user, but worry about
the user entering 0 or a negative num.
`int n; cin >> n; double *Z; Z = new double[n]; if( n <= 0 ) { cout << "Error!" << endl; Z = new double[10]; } Z[0] = 5;`

`int n; cin >> n; if( n <= 0 ) { cout << "Error!" << endl; double *Z = new double[10]; } else double *Z = new double[n]; Z[0] = 5;`

`int n; cin >> n; double *Z; if( n <= 0 ) { cout << "Error!" << endl; Z = new double[10]; } else Z = new double[n]; Z[0] = 5;`

compile-time error / possible run-time error / OK

If error, why?compile-time error / possible run-time error / OK

If error, why?compile-time error / possible run-time error / OK

If error, why? - [88pts]
Write a program that reads n points from standard input in the following format
N = n (x1,y1) (x2,y2) ... (xn,yn)

and writes out the x-coordinates of all n points on one line, and the y-coordinates of all n points on the following line. The output must be formatted like this:X := [x1, x2, ..., xn] Y := [y1, y2, ..., yn]

Here are some example runs. Your program must read input and produce output with the exact same formatting!~/$ ./hw N = 4 (0.5,1) (0.6,1.5) (0.7,2) (0.8,2.5) X := [0.5, 0.6, 0.7, 0.8] Y := [1, 1.5, 2, 2.5] ~/$./hw N = 1 (5,8) X := [5] Y := [8] ~/$./hw N = 10 (11,51) (12,52) (13,53) (14,54) (15,55) (16,56) (17,57) (18,58) (19,59) (20,60) X := [11, 12, 13, 14, 15, 16, 17, 18, 19, 20] Y := [51, 52, 53, 54, 55, 56, 57, 58, 59, 60]

**Note:**You may assume there is at least one point, i.e. you never get input like`N = 0`

.**Turn In**your source code and a screen capture of your program running on above example. (Hint: This is most easily done with two arrays.)