求逆序数(简单)

1.要求:

一组逆序数(a,b)=(2178  8712),

不仅数字顺序逆序,而且a*4=b

输入:正整数n

输出:小于正整数n的逆序数对

if __name__=="__main__":
    n=int(input())
    cnt=0
    result=[]
    if n<4:
        pass
    else:
        for i in range(1,int(n/4)+1):
            pre = str(i)
            after = 4*i
            after=str(after)
            if pre==after[::-1]:
                cnt+=1
                result.append([pre,after])
    if cnt>0:
        print(cnt)
        for ele in result:
            print(ele[0]," ",ele[1])
    else:
        print(cnt)
原文地址:https://www.cnblogs.com/sunupo/p/13510205.html