Python解决codeforces ---- 2

wsaer

　　

　　第一题 4A


A. Watermelon


time limit per test

1 second


memory limit per test

64 megabytes


input

standard input


output

standard output




One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showedwkilos.
They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.

Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The
boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.



Input


The first (and the only) input line contains integer numberw(1 ≤ w ≤ 100)
— the weight of the watermelon bought by the boys.



Output


PrintYES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; andNOin
the opposite case.



Sample test(s)


input


8



output


YES







Note


For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).













　　
　　题意：给定一个数w，问能否分成两部分并且每部份都是偶数
　　思路：w的值很小，暴力枚举两部分的值
　　代码：

n = int(raw_input())
isOk = False
for i in range(2,n):
if i%2 == 0 and (n-2)%2 == 0:
isOk = True
break
if isOk:
print "YES"
else:
print "NO"
　　

　　第二题 5A

A. Chat Server's Outgoing Traffic


time limit per test

1 second


memory limit per test

64 megabytes


input

standard input


output

standard output




Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop
and implemented a chat server that can process three types of commands:

• Include a person to the chat ('Add'command).
  
• Remove a person from the chat ('Remove'command).
  
• Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send'command).
  

Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands.

Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sendslbytes to each
participant of the chat, wherelis the length of the message.

As Polycarp has no time, he is asking for your help in solving this problem.



Input


Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following:

• +<name>for'Add'command.
  
• -<name>for'Remove'command.
  
• <sender_name>:<message_text>for'Send'command.
  

<name>and<sender_name>is a non-empty sequence of Latin letters
and digits.<message_text>can contain letters, digits and spaces, but can't start or end with a space.<message_text>can
be an empty line.

It is guaranteed, that input data are correct, i.e. there will be no'Add'command if person with such a name is already in the chat, there will
be no'Remove'command if there is no person with such a name in the chat etc.

All names are case-sensitive.



Output


Print a single number — answer to the problem.



Sample test(s)


input


+Mike
Mike:hello
+Kate
+Dmitry
-Dmitry
Kate:hi
-Kate



output


9



input


+Mike
-Mike
+Mike
Mike:Hi   I am here
-Mike
+Kate
-Kate



output


14





　　
　　题意：有三种命令，"+name"是添加一个人，"-name"是删除一个人，"name:message"是这个人发了message给所有人，问最后总的发送的字节数
　　思路：直接暴力求解，利用Python的list
　　代码：
　　

dict = []
sum = 0
# input
while True:
try:
str = raw_input()
except:
break
if str[0] == '+':
dict.append(str[1:])
elif str[0] == '-':
dict.remove(str[1:])
else:
length = len(str)  
for i in range(length):
if str == ':':
sum += (length-(i+1))*(len(dict))
break
print sum

　　

　　第三题 6A
　　


A. Triangle

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output



[size=1em]
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different
colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
[size=1em]
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a
degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.



Input

[size=1em]
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.



Output

[size=1em]
OutputTRIANGLEif it is possible to construct a non-degenerate triangle. OutputSEGMENTif
the first case cannot take place and it is possible to construct a degenerate triangle. OutputIMPOSSIBLEif it is impossible to construct any triangle. Remember
that you are to use three sticks. It is not allowed to break the sticks or use their partial length.



Sample test(s)

input

4 2 1 3



output

TRIANGLE



input

7 2 2 4



output

SEGMENT



input

3 5 9 1



output

IMPOSSIBLE



















　　题意：给定4条线段，问能否组成三角形，如果可以输出"TRIANGLE"，如果不能组成三角形但是会退化输出"SEGMENT"，否则输出"IMPOSSIBLE"
　　思路：直接暴力枚举
　　代码：
　　


sticks = raw_input().split()
# 判断能否组成三角形
def isOk(x , y , z):
if x+y > z and x+z > y and y+z > x:
return True
return False
# 判断是否会退化
def judge(x , y , z):
if x+y == z or x+z == y or y+z == x:
return True
return False
# 求ans
ans = "IMPOSSIBLE"
for i in range(4):
for j in range(i+1,4):
for k in range(j+1,4):
if isOk(int(sticks),int(sticks[j]),int(sticks[k])):
ans = "TRIANGLE"
if ans == "IMPOSSIBLE" and judge(int(sticks),int(sticks[j]),int(sticks[k])):
ans = "SEGMENT"
print ans