转移公式 就是 dp[n] = 1 + min([ dp[n - j**2] for j in xrange(1, int(n**0.5) + 1)]), 就是当前n对应的perfect square的数目减去一,即减去一个perfect square, 剩下的部分肯定也要是拥有最少perfect square的部分,并且这个要减去的perfect square,不能任意减,最多也就是减去自己的sqrt 这...
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