1.简单 AC
while True:
try:
n=int(input())
T=input()
flag=0
begin,end=len(T)-1,0
for i in range(len(T)):
if flag==0 and T[i]=='M':
flag=1
elif flag==1 and T[i]=='T':
begin=i+1
break
flag=0
for i in range(len(T)-1,-1,-1):
if flag==0 and T[i]=='T':
flag=1
elif flag==1 and T[i]=='M':
end=i-1
break
if begin<=end:
print (T[begin:end+1])
else:
print("")
except:
break 2.也比较简单 AC while True:
try:
n=int(input())
a=[]
for i in range(n):
a.append([int(s) for s in input().split()])
flag=[0]*n
res=[]
for i in range(n):
for j in a[i]:
if flag[j-1]==0:
flag[j-1]=1
res.append(j)
break
print(' '.join([str(r) for r in res]))
except:
break 3.这道题我是拓扑排序，看拓扑排序长度。一直是73% 可能是超时 while True:
try:
n,x,y=map(int,input().split())
edge=[]
for i in range(n-1):
uu,vv=map(int,input().split())
edge.append((uu,vv))
path={}
for uu,vv in edge:
if uu in path:
path[uu].append(vv)
else:
path[uu]=[vv]
if vv in path:
path[vv].append(uu)
else:
path[vv]=[uu]
print(path)
res=-1
visited=[0]*(n+1)
q=[x]
while q:
res+=1
q1=q
q=[]
for s in q1:
for pq1 in path[s]:
if visited[pq1]==0:
visited[pq1]=1
q.append(pq1)
print(res)
except:
break 4、AC。我的思路是先找出每个数字的左右边界，然后l从1到m判断l是否合理，如果合理的话，r就从m开始遍历到l，找到符合要求的数量就ok了。 while True:
try:
m,n=map(int,input().split())
seq=list(map(int,input().split()))
di={}
for i in range(n):
if seq[i] not in di:
di[seq[i]]=[i,i]
else:
max_i,min_i=di.get(seq[i])
di[seq[i]]=[max(max_i,i),min(min_i,i)]
res=0
for i in range(1,m+1):
mi,ma=0,n
flag=0
for j in range(1,i):
if j not in di:
continue
max_j,min_j=di.get(j)
if min_j>=mi:
mi=max_j
else:
flag=1
break
if flag==1:continue
res+=1
for j in range(m,i,-1):
if j not in di:
res+=1
continue
max_j,min_j=di.get(j)
if max_j<=ma and min_j>=mi:
res+=1
ma=min_j
else:
break
print(res)
except:
break 
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