Weave watermelons into No 1~ 10, the cutting number n goes into No.65438 +0/(2 n) and put it on the electronic scale.

If everything is normal, it should be m = (1/(21)+1(22)+1(23)+...+1(210).

The difference between m and actual weight m is1/(2k1)+1/(2k2)+.

(It is known from binary that k 1, k2 ... is unique)

K 1, k2 ... The corresponding number is a watermelon, which has a small weight.

If you have enough watermelons, it's better to use integers:

Watermelon racks are numbered 1~ 10, and (2 n) watermelons are taken from the nth rack and put on the electronic scale.

If everything is normal, it should be m = ((21)+(22)+(23)+...+(210)) * 5 (there should be 1024 watermelons in the tenth box).

The difference between m and the actual weight m is (2 k 1)+(2 k2)+ ...

(From the binary system, k 1, k2 ... are unique, that is, the binary representation of decimal numbers is unique. )

K 1, k2 ... The corresponding number is the watermelon box with less gold.