题解 | #在字符串中找出连续最长的数字串#

'''
解题思路：
原字符串前后加##，数字改1，其它改0，利用规则找01突变点，存储最长数字串
'''
while 1:
    try:

        L = input()
        L = '#'+L+'#'
        #print('L=',L)

        S=''
        for s in L:
            if s.isdigit():
                S += '1'
            else:
                S += '0'
        #print('S=',S)

        i_start = 0
        i_len = 0
        Out = []
        for i in range(1,len(S)):
            if S[i]=='1' and S[i-1]=='0':
                i_start = i
                #print('i_start=',i_start)
            elif S[i]=='0' and S[i-1]=='1':
                i_end = i
                #print('i_end=',i_end)
                if i_end - i_start == i_len:
                    Out.append(L[i_start:i_end])
                elif i_end - i_start > i_len:
                    i_len = i_end - i_start
                    Out = []
                    Out.append(L[i_start:i_end])

        #print(i_len)
        print(''.join(Out)+','+str(i_len))

    except:
        break

#---------------------------------------------------------------
'''
while True:
    try:
        a = input()
        maxLen, maxStrs, curLen, curStr = 0, [], 0, ""
        for i, v in enumerate(a):
            if v.isnumeric():
                curLen += 1
                curStr += v
                if curLen > maxLen:
                    maxLen = curLen
                    maxStrs = [curStr]
                elif curLen == maxLen:
                    maxStrs.append(curStr)
            else:
                curLen = 0
                curStr = ""
        print("".join(maxStrs) + "," + str(maxLen))
    except:
        break
'''