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Problem 719

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Number Splitting

We define an $S$-number to be a natural number, $n$, that is a perfect square and its square root can be obtained by splitting the decimal representation of $n$ into $2$ or more numbers then adding the numbers.

For example, $81$ is an $S$-number because $\sqrt{81} = 8+1$.
$6724$ is an $S$-number: $\sqrt{6724} = 6+72+4$.
$8281$ is an $S$-number: $\sqrt{8281} = 8+2+81 = 82+8+1$.
$9801$ is an $S$-number: $\sqrt{9801}=98+0+1$.

Further we define $T(N)$ to be the sum of all $S$-numbers $n\le N$. You are given $T(10^4) = 41333$.

Find $T(10^{12})$

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数字分割

若自然数$n$是个完全平方数，且其平方根可以表示为将$n$的十进制表示分成至少$2$部分再相加的和，则称其为$S$-数。

例如，$81$是个$S$-数，因为$\sqrt{81} = 8+1$。
$6724$是个$S$-数：$\sqrt{6724} = 6+72+4$。
$8281$是个$S$-数：$\sqrt{8281} = 8+2+81 = 82+8+1$。
$9801$是个$S$-数：$\sqrt{9801}=98+0+1$。

再记$T(N)$为所有满足$n\le N$的$S$-数之和。已知$T(10^4) = 41333$。

求$T(10^{12})$。

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