Today's problem I spent too much wrong on the format,I should be more careful.

Windows Message Queue

Description:
Message queue is the basic fundamental of windows system. For each process, the system maintains a message queue. If something happens to this process, such as mouse click, text change, the system will add a message to the queue. Meanwhile, the process will do a loop for getting message from the queue according to the priority value if it is not empty. Note that the less priority value means the higher priority. In this problem, you are asked to simulate the message queue for putting messages to and getting message from the message queue.
Input

There's only one test case in the input. Each line is a command, "GET" or "PUT", which means getting message or putting message. If the command is "PUT", there're one string means the message name and two integer means the parameter and priority followed by. There will be at most 60000 command. Note that one message can appear twice or more and if two messages have the same priority, the one comes first will be processed first.(i.e., FIFO for the same priority.) Process to the end-of-file.

Output

For each "GET" command, output the command getting from the message queue with the name and parameter in one line. If there's no message in the queue, output "EMPTY QUEUE!". There's no output for "PUT" command.

Sample Input

GET
PUT msg1 10 5
PUT msg2 10 4
GET
GET
GET

Sample Output

EMPTY QUEUE!
msg2 10
msg1 10
EMPTY QUEUE!

Problem solving:
Through a special queue with a struct and Operator overloading.The main question is understand this,oh my poor English.
Code:

#include <bits/stdc++.h>
using namespace std;
struct node
{
string s;
int    p, q, id;
} n[6005];
{
if (x.q == y.q)
return x.id > y.id;
return x.q > y.q;
}
priority_queue<node> que;
int main()
{
string s;
int    k = 0;
while (cin >> s)
{
if (s[0] == 'G')
{
if (que.empty())
puts("EMPTY QUEUE!");
else
{
cout << que.top().s << " " << que.top().p << endl;
que.pop();
}
}
if (s[0] == 'P')
{
cin >> n[0].s >> n[0].p >> n[0].q;
n[0].id = k;
k++;
que.push(n[0]);
}
}
}


Train Problem I

Description:
As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can't leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.

Input

The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.

Output

The output contains a string "No." if you can't exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.

Sample Input

3 123 321
3 123 312

Sample Output

Yes.
in
in
in
out
out
out
FINISH
No.
FINISH

For the first Sample Input, we let train 1 get in, then train 2 and train 3.
So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.
In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.
Now we can let train 3 leave.
But after that we can't let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.
So we output "No.".
Problem solving:
An easy problem.Use a stack to simulate this process and a queue to record 'in' or 'out'.Judge if we can pop or push.
Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
while (~scanf("%d", &n))
{
int           flag = 0;
string        a, b;
cin >> a >> b;
stack<char>   s;
queue<string> q;
for (int i = 0, j = 0; i < n && j <= n;)
{
if (s.empty() || s.top() != b[i])
{
if (j == n)
{
cout << "No.\nFINISH\n";
flag = 1;
break;
}
s.push(a[j]);
j++;
q.push("in");
}
else
{
s.pop();
q.push("out");
i++;
}
}
if (flag)
continue;
else
{
cout << "Yes.\n";
while (!q.empty())
{
cout << q.front() << endl;
q.pop();
}
}
cout << "FINISH\n";
}
}


Rails

Description:
There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.

The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, ..., N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, ..., aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

Input

The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, ..., N. The last line of the block contains just 0.
The last block consists of just one line containing 0.

Output

The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last null' block of the input.

Sample Input

5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0

Sample Output

Yes
No

Yes


Problem solving:
An easy stack question.
Code:

#include <iostream>
#include<cstdio>
#include<stack>
#include<cstring>
using namespace std;
int a[1005];
int main()
{
int n;
while (scanf("%d", &n) && n)
{
while (1)
{
int flag = 0;
for (int i = 0; i < n; i++)
{
cin >> a[i];
if (a[i] == 0)
{
flag = 1; break;
}
}
if (flag)
break;
stack<int> sta;
int        pos = 0;
for (int i = 1; i <= n; i++)
{
sta.push(i);
while (!sta.empty() && sta.top() == a[pos])
{
sta.pop();
pos++;
}
}
if (sta.empty())
puts("Yes");
else
puts("No");
}
puts("");
}
}


D - {A} + {B}

Description:

Input

Output

Sample Input

1 2
1
2 3
1 2
1
1 2

Sample Output

1 2 3
1 2

Problem solving:
An easy set question.
Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, m, a;
while (~scanf("%d %d", &n, &m))
{
set<int> s;
for (int i = 0; i < n + m; i++)
{
cin >> a;
s.insert(a);
}
for (set<int>::iterator it = s.begin(); it != s.end(); it++)
{
if (it == s.begin())
cout << *it;
else
cout << " " << *it;
}
cout << endl;
}
}


水果

Description:

Joe经营着一个不大的水果店.他认为生存之道就是经营最受顾客欢迎的水果.现在他想要一份水果销售情况的明细表,这样Joe就可以很容易掌握所有水果的销售情况了.
Input

Output

Sample Input

1
5
apple shandong 3
pineapple guangdong 1
sugarcane guangdong 1
pineapple guangdong 3
pineapple guangdong 1

Sample Output

guangdong
|----pineapple(5)
|----sugarcane(1)
shandong
|----apple(3)

Problem solving:
An unusual problem.We can ues a map in map.And then the iterator is a little different now.Just see the Code.Luckily,map can automatic sorting makes this problem is not so hard.
Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int    n, m, p;
string x, y;
cin >> n;
int xxxx=0;
while (n--)
{
if(xxxx!=0) puts("");
xxxx=1;
int                            i = 0;
map<string, map<string, int> > ma;
cin >> m;
while (m--)
{
cin >> x >> y >> p;
ma[y][x] += p;
}
map<string, int>::iterator iit;
for (map<string, map<string, int> >::iterator it = ma.begin(); it != ma.end(); it++)
{
cout << it->first << endl;
for (iit = it->second.begin(); iit != it->second.end(); iit++)
{
cout << "   |----" << iit->first << "(" << iit->second << ")" << endl;
}
}
}
}


Let the Balloon Rise

Description:
Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.
This year, they decide to leave this lovely job to you.
Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.
A test case with N = 0 terminates the input and this test case is not to be processed.

Output

For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

Sample Input

5
green
red
blue
red
red
3
pink
orange
pink
0

Sample Output

red
pink

Problem solving:
An easy map problem.
Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int    n;
string s;
while (cin >> n)
{
map<string, int> ma;
if (n == 0)
break;
while (n--)
{
cin >> s;
ma[s]++;
}
int    mid = 0;
string ans;
for (map<string, int>::iterator it = ma.begin(); it != ma.end(); it++)
{
if (it->second > mid)
{
ans = it->first;
mid = it->second;
}
}
cout << ans << endl;
}
}


不重复数字

Description:

Input

Output

Sample Input

2
11
1 2 18 3 3 19 2 3 6 5 4
6
1 2 3 4 5 6

Sample Output

1 2 18 3 19 6 5 4
1 2 3 4 5 6

Hint

Problem solving:
We can't use set here because set will automatic sort.I wanted to ues a flag array to know the number has appread or not.But always RE.So I choose a set to know if the number has already appeared.
Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int        n, a;
scanf("%d", &n);
set<int>   s;
int        mid = s.size();
queue<int> q;
while (n--)
{
scanf("%d", &a);
s.insert(a);
if (s.size() != mid)
{
q.push(a);
mid = s.size();
}
}

while (!q.empty())
{
cout << q.front() << " ";
q.pop();
}
puts("");
}
}


Andy's First Dictionary

Description:
Andy, 8, has a dream - he wants to produce his very own dictionary. This is not an easy task for him, as the number of words that he knows is, well, not quite enough. Instead of thinking up all the words himself, he has a briliant idea. From his bookshelf he would pick one of his favourite story books, from which he would copy out all the distinct words. By arranging the words in
alphabetical order, he is done! Of course, it is a really time-consuming job, and this is where a computer program is helpful.You are asked to write a program that lists all the different words in the input text. In this problem, a word is defined as a consecutive sequence of alphabets, in upper and/or lower case. Words with only one letter are also to be considered. Furthermore, your program must be CaSe InSeNsItIvE. For example, words like “Apple”, “apple” or “APPLE” must be considered the same.
Input

The input file is a text with no more than 5000 lines. An input line has at most 200 characters. Input
is terminated by EOF.

Output

Your output should give a list of different words that appears in the input text, one in a line. The
words should all be in lower case, sorted in alphabetical order. You can be sure that he number of
distinct words in the text does not exceed 5000.

Sample Input

Two blondes were going to Disneyland when they came to a fork in the
So they went home.

Sample Output

a
blondes
came
disneyland
fork
going
home
in
left
sign
so
the
they
to
two
went
were
when

Problem solving:

1. Turn uppercase letter to lower case letters.
2. Delete the char which is not a letter.

Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
string      s;
set<string> se;
while (cin >> s)
{
for (int i = 0; i < s.size(); i++)
{
if (s[i] >= 'A' && s[i] <= 'Z')
s[i] += 32;
else if (s[i] >= 'a' && s[i] <= 'z')
s[i] = s[i];
else
s[i] = ' ';
}
string       mi;
stringstream mid(s);
while (mid >> mi)
{
se.insert(mi);
}
}
for (set<string>::iterator it = se.begin(); it != se.end(); it++)
{
cout << *it << endl;
}
}


A and B and Compilation Errors

Description:
A and B are preparing themselves for programming contests.

B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.

Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.

However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.

Can you help B find out exactly what two errors he corrected?

Input

The first line of the input contains integer n (3 ≤ n ≤ 105) — the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers с1, с2, ..., сn - 2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.

Output

Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.

Examples
Input

5
1 5 8 123 7
123 7 5 1
5 1 7

Output

8
123

Input

6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5

Output

1
3

Note

In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.

Problem solving:
Through reding,you will kown the first line we output is the different number between the second line and the third line we input,and the second line we output is the different number between the third line and the fourth line we input.I choose a force way but easy to understand.
Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
int           n, a;
cin >> n;
map<int, int> ma;
map<int, int> ma1;
map<int, int> ma2;
for (int i = 0; i < n; i++)
{
cin >> a;
ma[a]++;
}
for (int i = 0; i < n - 1; i++)
{
cin >> a;
ma1[a]++;
}
for (int i = 0; i < n - 2; i++)
{
cin >> a;
ma2[a]++;
}
int x = 0, y = 0;
for (map<int, int>::iterator it = ma.begin(); it != ma.end(); it++)
{
int mid = it->first;
if (ma[mid] != ma1[mid])
cout << mid << endl;
}
for (map<int, int>::iterator it = ma1.begin(); it != ma1.end(); it++)
{
int mid = it->first;
if (ma2[mid] != ma1[mid])
cout << mid << endl;
}
return 0;
}


排列2

Description:
Ray又对数字的列产生了兴趣：

Input

Output

Sample Input

1 2 3 4
1 1 2 3
0 1 2 3
0 0 0 0

Sample Output

1234 1243 1324 1342 1423 1432
2134 2143 2314 2341 2413 2431
3124 3142 3214 3241 3412 3421
4123 4132 4213 4231 4312 4321

1123 1132 1213 1231 1312 1321
2113 2131 2311
3112 3121 3211

1023 1032 1203 1230 1302 1320
2013 2031 2103 2130 2301 2310
3012 3021 3102 3120 3201 3210


Problem solving:
An easy problem through 'next_permutation()'.The most disgusting place is the format.
Code:

#include <bits/stdc++.h>
using namespace std;
int a[5];
int main()
{
int mid, x = 0;
while (scanf("%d %d %d %d", &a[0], &a[1], &a[2], &a[3]) != EOF)
{
sort(a, a + 4);
if (a[0] == 0 && a[1] == 0 && a[2] == 0 && a[3] == 0)
break;
if (x)
cout << endl;
x = 1;
int flag = 1;
do
{
if (a[0] == 0)
continue;
if (flag)
{
cout << a[0] << a[1] << a[2] << a[3];
flag = 0;
}
else if (mid == a[0])
cout << " " << a[0] << a[1] << a[2] << a[3];
else
{
cout << endl;
cout << a[0] << a[1] << a[2] << a[3];
}
mid = a[0];
} while (next_permutation(a, a + 4));
puts("");
}
}


I have too much PE before I passed this problem.

Ignatius and the Princess II

Description:
Now our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill our pretty Princess. But now the BEelzebub has to beat our hero first. feng5166 says, "I have three question for you, if you can work them out, I will release the Princess, or you will be my dinner, too." Ignatius says confidently, "OK, at last, I will save the Princess."

"Now I will show you the first problem." feng5166 says, "Given a sequence of number 1 to N, we define that 1,2,3...N-1,N is the smallest sequence among all the sequence which can be composed with number 1 to N(each number can be and should be use only once in this problem). So it's easy to see the second smallest sequence is 1,2,3...N,N-1. Now I will give you two numbers, N and M. You should tell me the Mth smallest sequence which is composed with number 1 to N. It's easy, isn't is? Hahahahaha......"
Can you help Ignatius to solve this problem?
Input

The input contains several test cases. Each test case consists of two numbers, N and M(1<=N<=1000, 1<=M<=10000). You may assume that there is always a sequence satisfied the BEelzebub's demand. The input is terminated by the end of file.

Output

For each test case, you only have to output the sequence satisfied the BEelzebub's demand. When output a sequence, you should print a space between two numbers, but do not output any spaces after the last number.

Sample Input

6 4
11 8

Sample Output

1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10

Problem solving:
An easy problem through 'next_permutation()'.
Code:

#include <bits/stdc++.h>
using namespace std;
int a[1005];
int main()
{
int n, m;
while (~scanf("%d %d", &n, &m))
{
for (int i = 1; i <= n; i++)
{
a[i] = i;
}
int flag = 0;
do
{
flag++;
if (flag == m)
break;
} while (next_permutation(a + 1, a + n + 1));
for (int i = 1; i < n; i++)
cout << a[i] << " ";
cout << a[n] << endl;
}
}