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输入n后，使用for循环生成n*n阶方阵
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求解 n 的最小值，使得 g(n) ≥ f(n) 成立
问题：对于给定的函数 f(n) 和 g(n)，求解使 g(n) ≥ f(n) 恒成立的最小整数 n 值。示例：① f(n) = n，g(n) = nln(n)最小的 n 值为 2，即当 n > 2 时，g(n) ≥ f(n)。② f(n) = n^10，g(n) = 2n最小的 n 值为 约 10.
C source code:T((n) = T(n-1)+n,T(1) = 1
Solve the following recurrence relation by repeated substitution:T((n) = T(n-1)+n,T(1) = 1
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