SZU 2015 Winter Training Day#4

Problem(A15):TripleFree

Judge Info

Memory Limit: 32768KB

Case Time Limit: 10000MS

Time Limit: 10000MS

Judger: Normal

Description

Who can be the king in the kingdom of frogs?The one who can solve this problem in the shortest time. So, what’s theproblem? The problem named as Triple-Free- Or-Not. You are given a list ofinteger, you are asked to tell if the list is triple free or not.
Triple freemeans, no 3 integers in the list can form an increasing arithmetic sequence(asequence of numbers such that the difference of any two successive members is apositive constant). The old king wants to keep the position, he is asking yourhelp.

Task

You are given a list of integer numbers,write a program which can tell if it is triple free or not.

Input

The first line of input contains T(1 \leq T\leq 100), the number of test cases. The first line of each test case will havean integer number N(1 \leq N \leq 1, 000), the number of integers in the list.The second line will contain N integers separate by spaces,
all of them are inthe range of [-2^{31},2^{31}-1]\,.

Output

For each test case, print Triple-Free orTriple-NotFree in one line.

Sample Input

2

3

1 2 3

3

3 4 6

Sample Output

Triple-NotFree

Triple-Free

程序：
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int mod=10000;
const int MAXN=10000+5;

vector<int> point[MAXN];

int main()
{
int t,i,j,k,n,a[1005],line,pos,len;
long long temp;
memset(a,0,sizeof(a));

scanf("%d",&t);
while (t--)
{
for (i=0;i<MAXN;i++)
point[i].clear();

scanf("%d",&n);
for (i=0;i<n;i++)
{
scanf("%d",&a[i]);
point[abs(a[i])%mod].push_back(a[i]);
}

sort(a,a+n);
for (i=0;i<n;i++)
for (j=i+1;j<n;j++)
{
temp=2*a[j]-a[i];
line=abs(temp)%mod;
len=point[line].size();

if (!len) continue;
for (k=0;k<len;k++)
if (temp==point[line][k]) break;
if (k<len)
{
b2:
printf("Triple-NotFree\n");
goto b1;
}
}

printf("Triple-Free\n");

b1:;
}

return 0;
}

Problem(A41):Numbers

Judge Info

Memory Limit: 32768KB

Case Time Limit: 31000MS

Time Limit: 60000MS

Judger: Normal

Description

Let’s play a game about numbers. The rule ofthis game is quite simple. Only two kinds of commands in this game, one is’Add\, k\,’ command the other one is ’Ask\,

k\,’ command. You will have a list ofnumbers. At beginning the list is empty. When you got the command ’Add\, k’,firstly, you must check if the number k\, is in the list or not; secondly, ifthe number k\, is not in the list, add it to the list or do nothing.
When yougot the command ’Ask\, k’ you should tell if the number in the list or not.

Task

Please code a program for me, which can helpme to play this game.

Input

The first line of input contains T(1 \leq T\leq 20), the number of test cases. The first line of each test case will havean integer number N(1 \leq N \leq 1, 000, 000), the number of commands. N\,lines followed, each contains one command 'Add\, k' or 'Ask\,
k'(-\,2,000,000,000 \leq k \leq 2,000,000,000).

Hint for C++ users: Huge input, do not usecin.

Output

For each ’Ask\, k’ command, print ’YES\,’when the number k\, in the list or print ’NO\,’ in one line.

Sample Input

2

5

Add 2

Ask 3

Ask 2

Add 3

Ask 3

2

Add 1

Ask 2

Sample Output

NO

YES

YES

NO

程序：
#include <iostream>
#include <string>
#include <vector>
using namespace std;

const int mod=10000;
const int MAXN=10000+5;

vector<int> point[MAXN];

int main()
{
int n,temp,t1,line,i,len;
bool pos;
char s[5];

scanf("%d",&t1);

while (t1--)
{
for (i=0;i<MAXN;i++)
point[i].clear();

scanf("%d",&n);
while (n--)
{
scanf("%s",&s);
scanf("%d",&temp);

line=abs(temp%mod);
len=point[line].size();

pos=false;
for (i=0;i<len;i++)
if (temp==point[line][i])
{
pos=true;
break;
}

if (!strcmp(s,"Add"))
{
if (!pos)
point[line].push_back(temp);
}
else
{
if (pos)
printf("YES\n");
else
printf("NO\n");
}
}
}

return 0;
}

Problem(B40):Fengshui

Judge Info

Memory Limit: 32768KB

Case Time Limit: 10000MS

Time Limit: 30000MS

Judger: Normal

Description

Fengshui is an ancient subject in Chinesetradition. Someone considers it as science and someone criticizes it as blindfaith. Who knows! However, in modern days, everyone should respect culture fromour ancestor!

Fengshui focus on geography,environment andstaffs' position, all the theory come from a very old book named"YI". YI means change. Everything is always changing in the world.Fengshui wishes to guide changing, make life change to a better situation. Nowlet's
look at Fengshui's changing.

At first we must know about the traditionalfive elements system which composed by GOLD,WOOD,GROUND,WATER and FIRE.Everything in the world can be represented by one and only one element. Forexample, river is represented by WATER, hill is represented by GROUND.
Here, weonly consider the elements. In this system, once element can kill anotherelement, and one element can born anther element. Five elements compose as acircuit, as in Figure 1.

Every place has eight direction - east, west,north, south, northeast, northwest, southeast and southwest. Every directionhas a represented element. Now, our problem is about the elements at theseeight directions which form a Fengshui situation. Figure 2
is an example of oneFengshui situation.

But Fengshui situation can change! There'retwo change ways:

TURN: The whole situation turn clockwise onestep. Figure 3 shows the situation that situation in Figure 2 makes one TURNchange.

REBORN: Based on kill and born relation, onedirection's element can be killed by another direction's (at any other place)element in the situation, and then the killed element will born out as the newelement at its direction. Of course, kill and born are
all according as therelation of the system as in Figure 1. In situation of Figure 3, WATER in eastcan kill FIRE in southeast, then southeast place change to be GROUND, as inFigure 4.

Each change, no matter TURN or REBORN, constone step.

Now, there're two Fengshui situation, wewant to know is it possible that first one can change to the second one? And ifpossible, how many steps it need at least?

Input

There're several cases, the first line ofinput is the number of cases. Every case includes 6 lines, the first 3 linesindeicate the first Fengshui situation, the last 3 lines incicate the secondFengshui situation.

The format of one situation is as follow,there may be arbitrary blanks between adjacent directions.

northwest north northeast

west east

southwest south southeast

Output

For every case, output the number of theleast changing steps on a single line, if it is possible, or output -1.

Sample Input

2

GOLD WOOD WATER

WATER FIRE

WOOD GOLD GROUND

WATER GOLD WOOD

WOOD WATER

GOLD GROUND GROUND

WATER GROUND WOOD

GOLD FIRE

GOLD FIRE GROUND

GOLD FIRE FIRE

GOLD FIRE

WATER GROUND WOOD

Sample Output

2

14

Problem(B61):Games

Judge Info

Memory Limit: 32768KB

Case Time Limit: 10000MS

Time Limit: 10000MS

Judger: Normal

Description

Let's play a game like Problem A41 again.The rule of this game is quite simple. Only three kinds of commands in thisgame, 'Add \, k\,' , 'Ask \, k\,' and'Del \, k\,'. You will have a list of numbers. At beginning the list is empty.When you got the command
'Add \, k', firstly, you must check if the number 'k',is in the list or not; secondly, if the number 'k', is not in the list, add itto the list or do nothing. When you got the command 'Ask \, k' you should tellif the number in the list or not. When you got
the command 'Del\, k', firstly,you must check if the number 'k', is in the list or not; secondly, if thenumber 'k', is in the list, delete it from the list or do nothing.

Input

The first line of input contains T \,, thenumber of test cases. The first line of each test case will have an integernumber N(1 \leq N \leq 1, 000, 000), the number of commands. N \, linesfollowed, each contains one command 'Add \, k' , 'Ask \, k' and 'Del
\, k'. Forevery k \, fited in 32 \,-bit.

Output

For each 'Ask \, k' command, print 'YES \,'when the number k \, in the list or print 'NO \,' in one line.

Sample Input

2

5

Add 2

Ask 3

Ask 2

Del 2

Add 3

2

Del 2

Ask 1

Sample Output

NO

YES

NO

程序：
#include <iostream>
#include <string>
#include <vector>
using namespace std;

const int mod=10000;
const int MAXN=10000+5;

vector<long long> point[MAXN];

int main()
{
int n,temp,t1,line,i,len;
bool pos;
char s[5];

scanf("%d",&t1);

while (t1--)
{
for (i=0;i<MAXN;i++)
point[i].clear();

scanf("%d",&n);
while (n--)
{
scanf("%s",&s);
scanf("%d",&temp);

line=abs(temp%mod);
len=point[line].size();

pos=false;
for (i=0;i<len;i++)
if (temp==point[line][i])
{
pos=true;
break;
}

if (!strcmp(s,"Add"))
{
if (!pos)
point[line].push_back(temp);
}
else
if (!strcmp(s,"Ask"))
{
if (pos)
printf("YES\n");
else
printf("NO\n");
}
else
{
if (pos)
point[line][i]=3000000000;
}
}
}

return 0;
}