# 思路1 dp

* 时间复杂度 O(n \* sqrt(n))
* 空间复杂度 O(n)

For each i, it must be the sum of some number `(i - j*j)` and a perfect square number (j\*j). 即 i = (i - j \* j) + (j \* j)

dp\[n]表示最少使用的个数，则有(k为满足k^2<=n的最大的k)

转移方程

dp\[i] = min(dp\[i], dp\[i-j\*j]+1)

dp\[i] 可以取的最小值即为 1 + min(dp\[i-1], dp\[i-4], dp\[i-9] · · · )


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