cfdiv2_140

http://codeforces.com/contest/227/problem/A

A. Where do I Turn?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Trouble came from the overseas lands: a three-headed dragon Gorynych arrived. The dragon settled at point C and began to terrorize the residents of the surrounding villages.

A brave hero decided to put an end to the dragon. He moved from point A to fight with Gorynych. The hero rode from point A along a straight road and met point B on his way. The hero knows that in this land for every pair of roads it is true that they are either parallel to each other, or lie on a straight line, or are perpendicular to each other. He also knows well that points B and C are connected by a road. So the hero must either turn 90 degrees to the left or continue riding straight ahead or turn 90 degrees to the right. But he forgot where the point C is located.

Fortunately, a Brave Falcon flew right by. It can see all three points from the sky. The hero asked him what way to go to get to the dragon's lair.

If you have not got it, you are the falcon. Help the hero and tell him how to get him to point C: turn left, go straight or turn right.

At this moment the hero is believed to stand at point B, turning his back to point A.

Input

The first input line contains two space-separated integers x_a, y_a (|x_a|, |y_a| ≤ 10⁹) — the coordinates of point A. The second line contains the coordinates of point B in the same form, the third line contains the coordinates of point C.

It is guaranteed that all points are pairwise different. It is also guaranteed that either point B lies on segment AC, or angle ABC is right.

Output

Print a single line. If a hero must turn left, print "LEFT" (without the quotes); If he must go straight ahead, print "TOWARDS" (without the quotes); if he should turn right, print "RIGHT" (without the quotes).

Sample test(s)

Input

0 0

0 1

1 1

Output

RIGHT

Input

-1 -1

-3 -3

-4 -4

Output

TOWARDS

Input

-4 -6

-3 -7

-2 -6

Output

LEFT

Note

The picture to the first sample:

The red color shows points A, B and C. The blue arrow shows the hero's direction. The green color shows the hero's trajectory.

The picture to the second sample:

比赛的时候就是想不出来，用向量做了测试数据都过不了，表示题意没理解错，方法错了。。

题意：给3个点，A B C。。AB 表示一条有向射线，BC表示另一条有向射线，求ABC是否在同一条直线上||AB-BC向左成90度||AB——BC向右成90度。。。利用白书第五章最后一节说到的面积计算公式就行了。。

#include
#include
using namespace std;

int main()
{
    double  x1,y1,x2,y2,x0,y0;
    while(scanf("%lf%lf%lf%lf%lf%lf",&x0,&y0,&x1,&y1,&x2,&y2)!=EOF)
    {
        double area=x0*y1+x2*y0+x1*y2-x2*y1-x0*y2-x1*y0;
        if(area==0)
        {
            printf("TOWARDS\n");
        }
        else if(area>0)
        {
            printf("LEFT\n");
        }
        else
        {
            printf("RIGHT\n");
        }
    }
    return 0;
}

 

http://codeforces.com/contest/227/problem/B

B. Effective Approach

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.

According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.

Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to n) and ending with the n-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the n-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.

To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to n, and generated m queries of the form: find element with value b_i in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.

But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of elements in the array. The second line contains n distinct space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n) — the elements of array.

The third line contains integer m (1 ≤ m ≤ 10⁵) — the number of queries. The last line contains m space-separated integers b₁, b₂, ..., b_m (1 ≤ b_i ≤ n) — the search queries. Note that the queries can repeat.

Output

Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Sample test(s)

Input

2

1 2

1

1

Output

1 2

Input

2

2 1

1

1

Output

2 1

Input

3

3 1 2

3

1 2 3

Output

6 6

Note

In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).

In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).

 

 

B题：题意：有个数列，然后在输入若干个数，求这些数要通过几次比较才能找到（正着比较&&逆着比较）。。

#include
#include
using namespace std;
#define maxn 100100
long long int a[maxn];
long long int pos[maxn];
int main()
{
    int n,m;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            //scanf("%d",&a[i]);
            pos[a[i]]=i;
        }
        scanf("%d",&m);
        int b;
        long long int sum1=0,sum2=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&b);
            sum1+=pos[b];
            sum2+=n-pos[b]+1;
        }
        cout<<sum1<<" "<<sum2<<endl;
        //printf("%d %d\n",sum1,sum2);
    }
    return 0;
}