''' 解题思路: 构造一个素数判断函数isSu(n),搜索范围为 2 -> int(n**0.5+1) 两个数i、j相加搜索,搜索范围取一半 0 -> n//2+1 ''' def isSu(n): for i in range(2,int(n**0.5+1)): if n%i==0: return False return True while 1: try: pass n = int(input()) #print('n=',n) dis = n for i in range(n//2+1): #print(i,isSu(i)) j = n-i if isSu(i) and isSu(j) and abs(i-j)<dis: i_min = i j_min = j print(i_min) print(j_min) except: break