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函数题目！



=REDUCE(1,TAKE(AB:AB,A2),LAMBDA(x,y,LET(a,VSTACK(x,SEQUENCE(,ROWS(x),MAX(x)+ROWS(x),-1)),b,HSTACK(SEQUENCE(ROWS(a),,MAX(a)+ROWS(a),-1),a),c,VSTACK(SEQUENCE(,ROWS(b),MAX(b)+1),b),HSTACK(c,SEQUENCE(ROWS(c),,MAX(c)+1)))))


我擦 牛鼻


暂时没得思路


=LET(s,--CHOOSECOLS(SORT(HSTACK(SEQUENCE(A2*(A2+2)),DROP(REDUCE(1,SEQUENCE(A2),LAMBDA(x,y,HSTACK("",VSTACK(x,LET(a,2*y+1,b,2*y+1,MAKEARRAY(a,b,LAMBDA(r,c,IFNA(IFS(AND(r=1,c>1,c<a),r+c+a*a-a*2),"")))))))),1)),1,-1),SEQUENCE(,A2*2+2,2)),t,HSTACK(SEQUENCE(,A2+1)/0,1),z,--DROP(REDUCE(1,SEQUENCE(A2),LAMBDA(x,y,HSTACK("",VSTACK(x,LET(a,2*y+1,b,2*y+1,MAKEARRAY(a,b,LAMBDA(r,c,IFNA(IFS(c=1,-r+c+a*a-a*2+2,c=a,r+c+a*a-a*2,r=a,r-c+a*a-a*4+4),"")))))))),1),WRAPCOLS(TOCOL(VSTACK(s,t,z),3,1),A2*2+1))


纯数学方法，不难发现中心点上下左右均为系数恒定的二次函数，剩下的边边角角由这四个函数平移即可，因此构建四个数组互相替换
=LET(k,A2,MAKEARRAY(2*k+1,2*k+1,LAMBDA(r,c,IF(r<2*k+2-c,IF(r>c,4*(2+k-c)^2-9*(2+k-c)+7-r+k,4*(2+k-r)^2-7*(2+k-r)+3+c-k),IF(r>c,4*(r-k)^2-11*(r-k)+9+k-c,4*(c-k)^2-5*(c-k)+1+r-k)))))


=IFERROR((1-1*2)*IMAGINARY(IMDIV(IMDIV(COMPLEX(-(ROW(OFFSET(A1,,,A1*2+1))-(A1*2+1+1)/2),(COLUMN(OFFSET(A1,,,,A1*2+1))-(A1*2+1+1)/2)),IMCONJUGATE("1+i")),IMPOWER("-i",3-3*1-QUOTIENT(IMARGUMENT(IMDIV(COMPLEX(-(ROW(OFFSET(A1,,,A1*2+1))-(A1*2+1+1)/2),(COLUMN(OFFSET(A1,,,,A1*2+1))-(A1*2+1+1)/2)),IMCONJUGATE("1+i")))+PI()-0.001*1,PI()/2))))*2+((ABS((ROW(OFFSET(A1,,,A1*2+1))-(A1*2+1+1)/2)+(COLUMN(OFFSET(A1,,,,A1*2+1))-(A1*2+1+1)/2))+ABS((ROW(OFFSET(A1,,,A1*2+1))-(A1*2+1+1)/2)-(COLUMN(OFFSET(A1,,,,A1*2+1))-(A1*2+1+1)/2))))*ABS(3-3*1-MOD(QUOTIENT(IMARGUMENT(IMDIV(COMPLEX(-(ROW(OFFSET(A1,,,A1*2+1))-(A1*2+1+1)/2),(COLUMN(OFFSET(A1,,,,A1*2+1))-(A1*2+1+1)/2)),IMCONJUGATE("1+i")))+PI()-0.001*1,PI()/2),4))+((ABS((ROW(OFFSET(A1,,,A1*2+1))-(A1*2+1+1)/2)+(COLUMN(OFFSET(A1,,,,A1*2+1))-(A1*2+1+1)/2))+ABS((ROW(OFFSET(A1,,,A1*2+1))-(A1*2+1+1)/2)-(COLUMN(OFFSET(A1,,,,A1*2+1))-(A1*2+1+1)/2)))-1)^2,1)


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