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首先发现结果与需要改变的具体位置无关,只和需要改变的位置的个数有关,因此设f[i][j]表示选取了i个数字异或结果有j个1,只要分析接下来选择的数和这j个1有几个重合即可: 1. 三个数字全部重合,即$f[i][j+3]+=f[i-1][j]\cdot c(n-j,3)$ 2. 有两个数字重合,即$[i][j+1]+=f[i-1][j]\cdot c(n-j,2)\cdot j$ 3. 有一个数字重合,即$f[i][j-1]+=f[i-1][j]\cdot (n-j)\cdot c(j,2)$…
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