( 1949+2008)/2=[ 1978.5]= 1978

a4= 1993

a5= 1985

a6= 1989

a7= 1987

a8= 1988

a9= 1987

.......

a 1964= 1987

That's a good question. You can upgrade to how to stand in the right position among m people, so that the last shot is counted.

If this m is always divisible by 2 to 0, then stand at the end.

Otherwise, stand at the position where n is equal to the n power of 2 and is closest to m, for example, there are 10 people standing at the position of 2 3 = 8.