边界 f[1][0~2^k-1]=1
ans=[sigma(f[i][j])1<i<w/k,0<j<2^k-1]
如果w不能整除k,则ans还要加上 sigma(f[w/k+1][i]) 0<i<2^(w-w/k*k)-1
{w/k向下取整}
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