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问一个数列的求和公式a[1]=1,a[n]=a[n-1]+n的位数(a[n]=a[n-1]+int(lg(n))+1) 换句话说：a[1]到a[9]为公差为1的等差数列，a[9]到a[99]为公差为2的等差数列，a[99]到a[999]为公差为3的等差数列…… 这个数列的前n项和怎么求？ Followed by:
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