Three, (50 points) Find the number of non-negative integer solutions to the equation $\sum_{i=1}^{n}\left[\frac{x}{2^{k}}\right]=x-m(n \in$ $\left.\mathbf{N}_{+}, m \in \mathbf{Z}\right)$.
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\begin{array}{l}
\text { Three, Solution: Let } x=\left(a_{n} a_{n-1} \cdots a_{0}\right)_{2} \text {, where } a_{n} \in \mathbf{N}, a_{i} \in \\
\{0,1\}(0 \leqslant i \leqslant n-1) \text {. Then } \\
\sum_{k=0}^{n} a_{k} \cdot 2^{k}-m=\sum_{k=1}^{n}\left[\frac{x}{2^{k}}\right]=\left(a_{n} a_{n-1} \cdots a_{1}\right)_{2}+ \\
\left(a_{n} a_{n-1} \cdots a_{2}\right)_{2}+\cdots+\left(a_{n} a_{n-1}\right)_{2}+\left(a_{n}\right)_{2}=\sum_{k=0}^{n} a_{k} \cdot \\
\left(2^{k}-1\right) \text {, hence } \\
\sum_{k=0}^{n} a_{k}=m .
\end{array}
$$
\(\left(2^{k}-1\right)\), hence
(1) When \(m \leqslant -1\), by (1), the equation has no non-negative integer solutions for \(x\).
(2) When \(m=0\), by (1), we have \(a_{i}=0(0 \leqslant i \leqslant n)\), so the equation has 1 non-negative integer solution for \(x\), which is \(x=0\).
(3) When \(1 \leqslant m \leqslant n\), by (1), \(a_{n} \in \{m, m-1, \cdots, 0\}\). If \(a_{n}=m\), then \(a_{i}=0(0 \leqslant i \leqslant n-1)\), and the equation has 1 \(\left(\mathrm{C}_{n}^{0}\right)\) non-negative integer solution for \(x\); if \(a_{n}=m-1\), then one of \(a_{i}(0 \leqslant i \leqslant n-1)\) is 1 and the rest are 0, and the equation has \(\mathrm{C}_{n}^{1}\) non-negative integer solutions for \(x\); if \(a_{n}=m-2\), then two of \(a_{i}(0 \leqslant i \leqslant n-1)\) are 1 and the rest are 0, and the equation has \(\mathrm{C}_{n}^{2}\) non-negative integer solutions for \(x\); \(\cdots\); if \(a_{n}=0\), then \(m\) of \(a_{i}(0 \leqslant i \leqslant n-1)\) are 1 and the rest are 0, and the equation has \(\mathrm{C}_{n}^{m}\) non-negative integer solutions for \(x\). In summary,

the equation has \(\mathrm{C}_{n}^{0} + \mathrm{C}_{n}^{1} + \cdots + \mathrm{C}_{n}^{m} = \sum_{k=0}^{m} \mathrm{C}_{n}^{k}\) non-negative integer solutions for \(x\).
(4) When \(m > n\), by (1), \(a_{n} \in \{m, m-1, \cdots, m-n\}\). If \(a_{n}=m\), then \(a_{i}=0(0 \leqslant i \leqslant n-1)\), and the equation has 1 \(\left(\mathrm{C}_{n}^{0}\right)\) non-negative integer solution for \(x\); if \(a_{n}=m-1\), then one of \(a_{i}(0 \leqslant i \leqslant n-1)\) is 1 and the rest are 0, and the equation has \(\mathrm{C}_{n}^{1}\) non-negative integer solutions for \(x\); if \(a_{n}=m-2\), then two of \(a_{i}(0 \leqslant i \leqslant n-1)\) are 1 and the rest are 0, and the equation has \(\mathrm{C}_{n}^{2}\) non-negative integer solutions for \(x\); \(\cdots\); if \(a_{n}=m-n\), then all \(a_{i}(0 \leqslant i \leqslant n-1)\) are 1, and the equation has \(\mathrm{C}_{n}^{n}\) non-negative integer solutions for \(x\). In summary,

the equation has \(\mathrm{C}_{n}^{0} + \mathrm{C}_{n}^{1} + \cdots + \mathrm{C}_{n}^{n} = 2^{n}\) non-negative integer solutions for \(x\).
Combining (1), (2), (3), and (4), we have
When \(m \leqslant -1\), the equation has 0 non-negative integer solutions for \(x\);

When \(0 \leqslant m \leqslant n\), the equation has \(\sum_{k=0}^{m} \mathrm{C}_{n}^{k}\) non-negative integer solutions for \(x\);
When \(m > n\), the equation has \(2^{n}\) non-negative integer solutions for \(x\).
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