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- Update:
- 2008-10-13
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Description: cout < input the n: <<endl cin>> n int* p = new int [n+ 1] int** m = new int* [n+ 1] int** s = new int* [n+ 1] for (int i = 0 i <= n i++) (m [i] = new int [n+ 1] s [i] = new int [n+ 1]) cout << intput N* M (n+ 1 ): << endl for (i = 0 i <= n i++) cin>> p [i] int t
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