老刘1号

n=630
print([[x*100+1,(x+1)*100] for x in range(n//100)] + [[n//100*100+1,n]])
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老刘1号

回复 3# netdzb

print([[x*100+1,(x+1)*100] for x in range(n//100)] + (lambda n: [] if n % 100 == 0 else [[n//100*100+1,n]])(n))
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函数式编程感兴趣的话可以学一下racket