while True:
try:
l = []
s = input()
t = input()
for i in s:
for x in t:
if x == i:
l.append(i)
s1 = set(''.join(l))
s2 = set(s)
if s2 == s1:
print('true')
else:
print('false')
except:
break
while True:
try:
count = 0
s = input()
t = input()
for element in s:
if element in t:
count+=1
if count == len(s):
print('true')
else:
print('false')
except:
break这题的描述我还以为会把字符拆开,看到题解很多代码都是整个验证的,上当了
题解里的set方法真好,记一下
s1=input()
s2=input()
for one in s1:
if one not in s2:
print('false')
break
else:
print('true') short_s = input()
long_s = input()
#对短字符串与长字符串分别set操作并取交集,如果交集为短字符串set则表示短字符中所有字符都在长字符串中
if set(short_s)&set(long_s) == set(short_s):
print('true')
else:
print('false') ss=input()
ls=input()
count_ss=0
for i in ss:
if i in ls:
count_ss+=1
if count_ss == len(ss):
print('true')
else:
print('false')
s1 = set(list(input()))
s2 = set(list(input()))
if len(s1 - s2) == 0:
print('true')
else:
print('false') a=input()
b=input()
cc=0
for i in range(0, len(a)):
if a[i] in b:
cc=cc+1
if cc==len(a):
print('true')
else:
print('false')