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
inputCopy
5
black
black
white
white
black
outputCopy
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.
题意：
给你一个n，让你输入n个坐标，每输入一个的时候会给你这个坐标的颜色，让你最后找出一条直线把所有的点分成两部分，相同颜色的在一边。
题解：
看到这个30的数据范围就知道要搞事情了，我们可以设一个l,r，l=0,r=1<<29，因为第一个点在0，所以只剩下29个点了，不会超1e9.第二个点的位置就是l+r>>1,颜色如果和第一个点一样，那么就继续往右，否则往左，想想就知道这是为什么了，最后的线就可以穿过中间的随便一条线。有一个情况就是l==0的时候，我们线就要设成1 1 和2 3，这样就没问题咯。
#include<bits/stdc++.h>
using namespace std;
int n,line;
int main()
{
scanf("%d",&n);
n-=1;
int l=0,r=1<<29;
line=1;
printf("0 0\n");
string s;
int flag;
fflush(stdout);
cin>>s;
flag=s[0];
int turn;
while(n--)
{
int mid=l+r>>1;
printf("%d 0\n",mid);
fflush(stdout);
cin>>s;
if(s[0]==flag)
l=mid,turn=1;
else
r=mid,turn=0;
}
if(l==0)
{
printf("1 1 2 3\n");
}
else
{
if(turn)
printf("%d 1 %d 3",l,l-1);
else
printf("%d 1 %d 3",r,r+1);
}
return 0;
}