我想进行Python的时间加减运算

import time
import datetime
t=input()
t=time.strptime(t,'%H:%M:%S')
H,M,S=t[3:6]
seconds=eval(input())
seconds=seconds%(24*3600)
hour=seconds//3600
seconds%=3600
minutes=seconds//60
seconds%=60
t1=datetime.timedelta(hours=H+hour,minutes=M+minutes,seconds=seconds+S)
print(t1)
这个是我写的代码   

运行结果如下

我想删除前面的1 days怎么办啊

是想删除前面的days, 数还需要添加到后面的时分秒吗?

试下这个:

import time
import datetime
t=input()
t=time.strptime(t,'%H:%M:%S')
H,M,S=t[3:6]
seconds=eval(input())
seconds=seconds%(24*3600)
hour=seconds//3600
seconds%=3600
minutes=seconds//60
seconds%=60
t1=datetime.timedelta(hours=H+hour,minutes=M+minutes,seconds=seconds+S)
print(t1)

from datetime import timedelta
def format_timedelta(td):
    minutes, seconds = divmod(td.seconds + td.days * 86400, 60)
    hours, minutes = divmod(minutes, 60)
    return '{:d}:{:02d}:{:02d}'.format(hours, minutes, seconds)
str_t1 = str(t1)
format_timedelta(timedelta(days=int(str_t1.split(' ')[0]), 
                           hours=int(str_t1.split(' ')[2].split(':')[0]), 
                           minutes=int(str_t1.split(':')[1]), 
                           seconds=int(str_t1.split(':')[2])))

要删除前面日期的话，就再用一次timedelta函数去减。

import time
import datetime
t = input()
t = time.strptime(t, '%H:%M:%S')
H, M, S = t[3:6]
seconds = eval(input())
seconds = seconds % (24*3600)
hour = seconds//3600
seconds %= 3600
minutes = seconds//60
seconds %= 60
t1 = datetime.timedelta(hours=H+hour, minutes=M+minutes, seconds=seconds+S)
days=t1.days
t1=t1-datetime.timedelta(days=days)
print(t1)