# 3 - 7
n =int(input())
lst =list(map(int,input().split()))max= lst[0]
index =0for i inrange(1,len(lst)):if(max< lst[i]):max= lst[i]
index = i
print('{} {}'.format(max,index))# 3 - 9"""
输入一个以#结束的字符串，本题要求滤去所有的非十六进制字符（不分大小写），组成一个新的表示十六进制数字的字符串，
然后将其转换为十进制数后输出。如果在第一个十六进制字符之前存在字符“-”，则代表该数是负数。
silu :读取整行字符串，然后到“#”截止，取出十六进制部分，用flag标记是否为负数。
""""""
class int(x, base=10) 默认十进制
int() 函数用于将一个字符串或数字转换为整型。
"""# 法一：
s =input()
fig ='abcdefABCDEF'sum=0
lis1=''for i in s:if(i in fig)or(i.isdigit())or(i=='-'):
lis1=lis1+i
if i=='#':breakif(lis1[0]=='-'):
lis1='-'+lis1.replace('-','')else:
lis1=lis1.replace('-','')print(int(lis1,16))# s=input()# st=""# for i in s:# if(i=='#'):# break# st=st+i# t=""# flag=0# f=0# sum=0# for i in st:# if(i=='-' and flag==0):# f=1# if(i>='0' and i<='9')or(i>='a' and i<='f')or(i>='A' and i<='F'):# t=t+i# flag=1# if(f): # 是不是负数# for i in t:# if (i >= '0' and i <= '9'):# sum = sum * 16 + ord(i) - 48 # ord() 返回ASCII值# elif (i >= 'a' and i <= 'f'):# sum=sum * 16+ord(i)-97# elif (i>= 'A' and i <= 'F'):# sum=sum * 16+ord(i)-65# print(sum*-1)# else:# for i in t:# if (i >= '0' and i <= '9'):# sum = sum * 16 + ord(i) - 48# elif (i >= 'a' and i <= 'f'):# sum=sum * 16+ord(i)-97# elif (i>= 'A' and i <= 'F'):# sum=sum * 16+ord(i)-65# print(sum)# 3 - 10
cnt =0
a ='AEIOU'
lst =input()for i in lst :if( i>='A'and i<='Z'and i notin a ):
cnt +=1print(cnt)# 3 - 11
lst =list(map(str,input().split()))
lst.sort()print("After sorted:")for i inrange(0,len(lst)):print(lst[i])"""
改进：一行
print("After sorted:", *sorted(list(input().split())), sep="\n")
"""# 3 -13
a =input()
b =[]for n in a :if"A"<= n <="Z":
b.append(chr(155-ord(n)))else:
b.append(n)print("".join(b))# 3 -14
s =input()
s=s.replace('#','')
t=""for n in s:if(n.islower()):
t=t+n.upper()elif(n.isupper()):
t=t+n.lower()else:
t=t+n
print(t)# 3 -15# 统计单词个数 这可比c简单多了...
cnt=0for i ininput().split():
cnt=cnt+1print(cnt)# replace方法用来删除字符# 3 - 17
s1 =input().strip()
s2 =input().strip()
s1 = s1.replace(s2.upper(),'').replace(s2.lower(),'')print("result: {}".format(s1))# 3 - 18
s=input()
new=""
count=0for i in s:if(i.islower()or i.isupper()):if(i.upper()notin new)and(i.lower()notin new):
new+=i
count+=1if count<10:print("not found")else:print(new[:10])"""
法二：
a=input()
c=''
b=[]
for i in a:
if i.isupper() or i.islower():
c+=i
for i in c:
if b.count(i.lower())==0 and b.count(i.upper())==0:
b.append(i)
print(''.join(b[:10]) if len(b)>=10 else 'not found')
"""# 3 - 19# 最长字符串
n=int(input())max=0
maxstr=""for i inrange(0,n):
s=input()if(max<len(s)):max=len(s)
maxstr=s
print('The longest is: {:s}'.format(maxstr))# 3 - 20 逆序三位很简单 多位的话就除去前面的0即可（切片操作）
n=input()
a=int(n[0])
b=int(n[1])
c=int(n[2])print(a+b*10+c*100)# 3 - 21 回文数
s =input()
a =reversed(list(s))# 返回一个迭代器iflist(a)==list(s):print(s)print("Yes")else:print(s)print("No")# 3 - 22
s =input()
t=""
flag=0for i in s:if(i.isupper()):
t=t+i
flag=1if(flag==0):print("Not Found")else:
lst=list(t)
res=list(set(lst))
res.sort(key=lst.index)# 因为set是一种会自动排序的数据结构，会打乱原来的顺序，需要用res.sort(key=lst.index)去恢复原来的顺序print("".join(res))# 4 - 4import math
defisPrime(n):if n <=1:returnFalsefor i inrange(2,int(math.sqrt(n))+1):# 根号范围足够判断 防止超时if n % i ==0:returnFalsereturnTrue
x=int(input())for y inrange(2,x//2+1):
z = x - y
if(isPrime(y)==1and isPrime(z)==1):print('{:d} = {:d} + {:d}'.format(x, y, z))break# 4 - 6# 求前n个斐波那契数deffib(n):
a, b =0,1for i inrange(n +1):
a, b = b, a + b
return a
n=int(input())if(n>0):for i inrange(0,n):print('{:11d}'.format(fib(i)),end="")if((i+1)%5==0):print("")else:print("Invalid.")# 4 - 7 注意考虑0
n=int(input())if(n==0):print('average = 0.0')print('count = 0')
exit(0)
lst =list(map(int,input().split()))sum=sum(lst)
aver=sum/n
new_lst =[x for x in lst if x >=60]
cnt =len(new_lst)print('average = {:.1f}'.format(aver))print('count = {:d}'.format(cnt))# 4 - 10defgcd(a,b):if a%b ==0:return b
else:return gcd(b,a%b)
n , m =input().split()
n =int(n)
m =int(m)print('{:d} {:d}'.format(gcd(n,m),n*m//gcd(n,m)))# print('{:d} {:.0f}'.format(gcd(n,m),n*m/gcd(n,m)))""" 注意除法是带小数的，可以做上述两种处理 """# 输出小于n的最大斐波那契数
n =int(input())
a,b =1,1while b<=n:
a,b = b,a + b
print(b)# 4 - 13
error =float(input())
sum1 =1
sum2 =2
n =1
term =1while(sum2 - sum1)>= error:
term *=1/n
sum1 += term
sum2 += term *1/(n+1)
n = n +1print('{:.6f}'.format(sum2))# 4 - 14""" 循环输入的方式解决了input遇到回车的问题 """
s=[]
count=0;letters=0;space=0;digit=0;others=0whileTrue:
b=list(input())
count=count+1
s.extend(b)iflen(s)+count>10:
count=count-1breakfor c in s:if c.isalpha():
letters+=1elif c.isspace():
space+=1elif c.isdigit():
digit+=1else:
others+=1print("letter = {}, blank = {}, digit = {}, other = {}".format(letters,space+count,digit,others))# 4 - 15 换硬币
n=int(input())
cnt=0for fen5 inrange(n//5,0,-1):for fen2 inrange(n//2,0,-1):for fen1 inrange(n,0,-1):if(fen5 *5+fen2 *2+fen1 == n):print('fen5:{:d}, fen2:{:d}, fen1:{:d}, total:{:d}'.format(fen5, fen2, fen1, fen1+fen2+fen5))
cnt=cnt+1print('count = {:d}'.format(cnt))# 4 - 17 水仙花数import math
n=int(input())for i inrange(int(math.pow(10,n-1)),int(math.pow(10,n))):sum=0
j=i
while(j>=1):sum=sum+math.pow(j%10,n)
j=j//10if(sum==i):print('{:d}'.format(i))# 4 - 18 猴子选大王 约瑟夫环问题
n=int(input())if n ==1:print(1)else:
monkey=[i for i inrange(1,n+1)]whilelen(monkey)>=3:
monkey.pop(2)
monkey.append(monkey.pop(0))
monkey.append(monkey.pop(0))print(monkey[1])# 最后剩下的两者中后报数的"""
法二：
def yuesefu(nums, step, stay):
#参数 nums：人数，step: 数到几的步数，stay: 最后留下多少人
lists = list(range(1, nums+1))
check = 0
while len(lists) > stay:
for i in lists[:]:
check += 1
if check == step:
check = 0
lists.remove(i)
print("{}号下船了".format(i))
return lists
stays = yuesefu(11, 3, 1)
print("最后留下的人：", stays)
"""# 4 - 19 矩阵运算"""
给定一个n×n的方阵，本题要求计算该矩阵除副对角线、最后一列和最后一行以外的所有元素之和。副对角线为从矩阵的右上角至左下角的连线
输入第一行给出正整数n（1<n≤10）；随后n行，每行给出n个整数，其间以空格分隔。
思路：
副对角线为j+k=n-1。二维呈现是一维，因为python没有数组，只有列表 所以索引应该是
这种结构：list = [ [],[],[],]
"""
n=int(input())
a=[]sum=0for i inrange(n):
s=input()
a.append([int(n)for n in s.split()])for j inrange(n):# 循环遍历for k inrange(n):if j!=n-1and k!=n-1and j+k!=n-1:sum+=a[j][k]print(sum)"""
法二：
n=int(input())
s=0
for i in range(1,n+1):
m=list(map(int,input().split())) # m 每次用完就废了
if i==1:
m.pop(-i)
s=s+sum(m)
elif i==n:
s=s # break
else:
m.pop(-i)
m.pop(-1)
s=s+sum(m)
print(s)
"""# 4 - 20
m,n=map(int,input().split())
a=[]for i inrange(m):
s=input()
a.append([int(n)for n in s.split()])# 每次加一个列表 记住需要【】或者list（） 否则列表生成式返回的是generator objectfor j inrange(m):sum=0for k inrange(n):sum+=a[j][k]print(sum)# 4 - 21 判断上三角矩阵
T =int(input())for i inrange(T):
a =[]
flag =0
n =int(input())for j inrange(n):
s =input()
a.append([int(n)for n in s.split()])#上式另一种写法可以是# s=list(map(int,input().split()))# a.append(s)for j inrange(n):for k inrange(n):if j > k:if a[j][k]!=0:
flag =1if flag ==0:print('YES')else:print('NO')# 4 - 22 找鞍点
n =int(input())
a =[]
f =0defchoice(a,i,j,n):
flag =0
s =sorted(a[i])
m =[a[k][j]for k inrange(n)]
m =sorted(m)if(a[i][j]== s[n-1]and a[i][j]== m[0]):
flag =1return flag
for i inrange(n):
s =[int(j)for j ininput().split()]
a.append(s)for i inrange(n):for j inrange(n):if(choice(a,i,j,n)==1):print('{} {}'.format(i,j))
f =1breakif(f ==1):breakif(f ==0):print('NONE')""" 法二：
n = int(input())
a = []
count = 0
count1 = 0
for i in range(n):
s = input()
a.append([int(n) for n in s.split()])
for j in range(n):
if count1 == n and count == n:
break
for k in range(n):
for k1 in range(n):
if a[j][k] >= a[j][k1]:
count += 1
if count == n: # 如果是行最大再判断列
for j1 in range(n):
if a[j][k] <= a[j1][k]:
count1 += 1
if count1 == n: # 如果是列最小就输出
print("{} {}".format(j, k))
break
count1 = 0
count = 0
if count1 != n and count != n:
print("NONE")
"""# 4 - 23# 矩阵局部最大值
n, m=input().split()
n, m =int(n),int(m)
a =[]sum=0
count=0
r=0# sum = 0;count=0;r=0 可以用分号去分割for i inrange(n):
s =input()
a.append([int(n)for n in s.split()])for i inrange(1,n-1):for j inrange(1,m-1):if(a[i][j]>a[i-1][j]and a[i][j]>a[i][j-1]and a[i][j]>a[i+1][j]and a[i][j]>a[i][j+1]):
r=1print('{:d} {:d} {:d}'.format(a[i][j],i+1,j+1))if(r==0):print('None {:d} {:d}'.format(i+2,j+2))# 4 - 24 九九乘法
n=int(input())for i inrange(1,n+1):for j inrange(1,i+1):print('{:d}*{:d}={:<4d}'.format(j,i,i*j),end="")print("")"""
一句话：
print('\n'.join([' '.join('%sx%s=%-2s'%(x,y,x*y) for x in xrange(1,y+1)) for y in xrange(1,10)]))
"""# 4 - 25
n =int(input())
cnt =65for i inrange(n):for k inrange(n):print('{:c} '.format(cnt),end="")
cnt +=1
n = n-1print("")# 4 - 26 Q: 1!+3!+5!...n!
n =int(input())
s =0for i inrange(1,n+1,2):
k =1for j inrange(1,i+1):
k *= j
s += k
print('n={},s={}'.format(n,s))# 4 - 28
s=list(map(int,input().split()))for i inrange(0,3):print("{:>4} {:>4} {:>4}".format(s[i],s[i+3],s[i+6]))# 4 - 29 找出不是两个数组共有的元素 (20 分)# 关键问题其实在于如何从输入中读取数据
c=[]
k =0
flag =1
a =list(map(int,input().split()))
b =list(map(int,input().split()))
m = a[0]
n = b[0]
a = a[1:]
b = b[1:]for i inrange(0, m):for j inrange(0,n):if(int(a[i])==int(b[j])):
flag=0breakif(flag==1):
c.append(a[i])
k=k+1
flag=1for i inrange(0,n):for j inrange(0,m):if(int(b[i])==int(a[j])):
flag=0breakif(flag==1):
c.append(b[i])
k=k+1
flag=1
res =list(set(c))
res.sort(key = c.index)print(res[0],end="")for i inrange(1,len(res)):print(' {:d}'.format(res[i]),end="")""" 利用python特性
s=list(map(int,input().split()))
t=list(map(int,input().split()))
lens=s.pop(0) # len 没用上
lent=t.pop(0)
l=[]
for i in s:
if i not in t and i not in l:
l.append(i)
for i in t:
if i not in s and i not in l:
l.append(i)
print(*l,sep=' ')
# 也可以用
temp = [str(i) for i in temp]
temp = ' '.join(temp)
print(temp)
"""# 4 - 30 找完数"""
这个是自己的方法，但是运行超时，注意是因为分解因数到根号i即可
"""
m,n =input().split()
m,n =eval(m),eval(n)
a =[]for i inrange(m,n+1):for j inrange(2,i):if(i % j ==0):
a.append(j)if(i ==sum(a)+1):print('{} = 1 '.format(i),end="")for k in a:print('+ {} '.format(k),end="")print("")
a=[]"""
不超时的：
import math
l = [1]
flag = 0
m,n=input().split()
m=int(m)
n=int(n)
for i in range (m,n+1):
# l.append(1)
for a in range (2,int(math.sqrt(i)+1)):
if i%a ==0:
l.append(a)
if a*a!=i:
l.append(i//a)
if sum(l)==i:
print ('{:d} = '.format(i),end="")
l.sort()
print(" + ".join(['%s' %id for id in l]))
# %d也行 不加【】也行 但是不能直接写l 即不能直接是数字列表
flag=1
l = [1]
if(flag == 0):
print("None")"""# 5 - 1 字典代替if else判断
days ={1:"Mon",2:"Tue",3:"Wed",4:"Thu",5:"Fri",6:"Sat",7:"Sun"}
num =int(input())print(days[num])# 5 - 3 实现四则运算dict={"+":"x+y","-":"x-y","*":"x*y","/":"x/y"}
x =float(input())
op =input()
y =float(input())try:print("{:.2f}".format(eval(dict[op])))except:# 0用异常处理print("divided by zero")"""
字典+eval（）的妙用！
res={"+":"x+y","-":"x-y","*":"x*y","/":"x/y if y!=0 else 'divided by zero'"}
# 除法部分为三元表达式
x = int(input())
operation = input()
y = int(input())
# print(res[operation]) 除零返回字符串，其余返回数字类型
r = eval(res[operation])
if type(r)!=str:
print(format(r,'.2f')) # format使用
else:
print(r)
"""# 5 - 4
l=list(map(int,input().split(',')))
s=[i for i inrange(6,11)if i notin l]print(*s,sep=' ')"""
这种写法还是想强调join是对于字符串列表的
str1 = set(input().split(","))
ls = []
for i in range(6,11):
if str(i) not in str1:
ls.append(str(i)) # str不可少
ll = ' '.join(ls)
print(ll)
"""# 5 - 5:
s=list(input())
dic={}for i in s:
dic[i]=dic.get(i,0)+1try:
m=input()print(dic[m])except:print(0)"""
查数的都可以尝试用count
str1 = input()
char1 = input()
print(str1.count(char1))
"""# 5 - 6
dic ={}
n =int(input())
lst =list(map(int,input().split()))for i in lst :
dic[i]= dic.get(i,0)+1# 善用字典的get方法！for j insorted(dic):# 排序print("{}:{}".format(j, dic[j]))""" 法二：未使用get方法就显得比较繁琐
n=int(input())
lst = list(map(int,input().split()))
d = {}
for x in range(0,n):
if not lst[x] in d:
d[lst[x]] = 1
else:
d[lst[x]] = d[lst[x]] + 1
for k in sorted(d.keys()):
print(k,end=':')
print(d[k])
"""
