很明显,求方程x*2^m=L (mod n+1)的解(1<=x<=n) var n,m,l,x,y,p:int64; function quick(x:int64):int64; var j:int64; begin j:=; quick:=; do begin = then quick:=quick*j mod p; x:=x ; j:=j*j mod p; end; end; procedure exgcd(a,b:int64;var x,y:int64); var xx,yy:int64…