CodeForces - 1064E Dwarves, Hats and Extrasensory Abilities(二分)

E. Dwarves, Hats and Extrasensory Abilities

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

This is an interactive problem.

In good old times dwarves tried to develop extrasensory abilities:

  • Exactly n dwarves entered completely dark cave.
  • Each dwarf received a hat — white or black. While in cave, none of the dwarves was able to see either his own hat or hats of other Dwarves.
  • Dwarves went out of the cave to the meadow and sat at an arbitrary place one after the other. When a dwarf leaves the cave, he sees the colors of all hats of all dwarves that are seating on the meadow (i.e. left the cave before him). However, he is not able to see the color of his own hat and none of the dwarves can give him this information.
  • The task for dwarves was to got diverged into two parts — one with dwarves with white hats and one with black hats.

After many centuries, dwarves finally managed to select the right place on the meadow without error. Will you be able to repeat their success?

You are asked to successively name n different integer points on the plane. After naming each new point you will be given its color — black or white. Your task is to ensure that the named points can be split by a line in such a way that all points of one color lie on the same side from the line and points of different colors lie on different sides. Moreover, no points can belong to the line. Also, you need to report any such line at the end of the process.

In this problem, the interactor is adaptive — the colors of the points in the tests are not fixed beforehand and the jury program can select them arbitrarily, in particular, depending on your program output.

Interaction

The first line of the standard input stream contains an integer n (1 ≤ n ≤ 30) — the number of points your program should name.

Then n times your program must print two integer coordinates x and y (0 ≤ x ≤ 109, 0 ≤ y ≤ 109). All points you print must be distinct.

In response to each coordinate pair your program will receive the string "black", if the point is black, or "white", if the point is white.

When all n points are processed, you need to print four integers x1, y1, x2 and y2 (0 ≤ x1, y1 ≤ 109, 0 ≤ x2, y2 ≤ 109) — coordinates of points (x1, y1) and (x2, y2), which form a line, which separates n points into black and white. Points (x1, y1) and (x2, y2) should not coincide.

Hacks

To hack solution use the following format. The first line must contain word "hack", the second line should contain the number n and the last line should contain the sequence of 0 and 1 — colors of points, which will be reported to the solution. Unlike the jury tests, colors of points in hacks are always fixed in advance. Of course, the hacked solution wouldn't be able to get the information about the colors in advance.

For example, the hack corresponding to sample test will look like this:

hack
5
0 0 1 1 0

Example

input

Copy

5

black

black

white

white

black

output

Copy

0 0

3 1

2 3

4 4

0 2

1 3 4 1

Note

In the sample input and output values are aligned only for simplicity of interpreting them chronologically. In real interaction no "extra" line breaks should appear.

The following picture illustrates the first test.

CodeForces - 1064E Dwarves, Hats and Extrasensory Abilities(二分)

 

题意:这是一个互动题,简单地说就是你给对方很多个点,对方会把你这些点染色,最后你要输出一条直线,把黑白两重点分开。

 

 

解题思路:二分思想,我们在 X轴二分, 我们先给出0,0这个点

假设给出了白色,那么我们 再给出  500000000,0 这个点

如果也是白色,那么就往右区间给点,如果是黑色就往左区间给点

这样子最后肯定成立。

题目卡了一点点精度,所以我们最后的中线要斜着取。因为斜着取可能出现负数,所以我们在Y=10000这个轴二分好了

CodeForces - 1064E Dwarves, Hats and Extrasensory Abilities(二分)

 

 

#include<bits/stdc++.h>
using namespace std;
const int MAXN=100005;
typedef long long ll;


int main(){

    int N;
    cin>>N;
    string col;
    int L=0,R=1000000000;
    int m=0;
    cout<<0<<" "<<1000000<<endl;
    string fir;
    for(int i=0;i<N;i++){
        cin>>col;
        if(i==0)
            fir=col;

        if(fir==col)
            L=m;
        else
            R=m;


        m=(L+R)/2;

        if(i==N-1){
            cout<<L<<" "<<1000001<<" "<<R<<" "<<1<<endl;
        }
        else
            cout<<m<<" "<<1000000<<endl;


    }

    return 0;
}