1079 延迟的回文数 (20 分)python

首先要判断输入的数是否为回文如(1,11,121,123321,1234321),如果是则直接输出,不是再进入循环。
a = input()
count = 0
flag = 0
if a == a[::-1]:
    print("%s is a palindromic number." % a)
    flag =1
if flag==0:
    while 1:
        if count == 10:
            print("Not found in 10 iterations.")
            break
        count = count+1
        a1 = a[::-1]
        b = str(int(a) + int(a1))
        print("%d + %d = %d"%(int(a),int(a1),int(b)))
        a = b
        if a == a[::-1]:
            print("%s is a palindromic number." % a)
            break
        else:
            continue

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转载自blog.csdn.net/weixin_41775301/article/details/87892558