題意

交互題, 本來應該是在平面上進行的.
實際上換成一條直線就可以, 其實換成在平面上更復雜一些.

Solution

假設\(l\)點是黑點, \(r\)處是白點, 那麽就把下一個點的位置放置在\(l + r / 2\)處, 然後遞歸處理.

Code

#include <ctype.h>
#include <stdio.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 39;

struct Node {
    int color, position;
    bool operator < (const Node &x) const {
        return color == x.color ? position < x.position : color < x.color;
    }
} p[N];
const int XXX = 4323;

bool Check(int position, int now) {
    string s;
    p[now].position = position;
    printf("%d %d\n", XXX, position);
    fflush(stdout);
    cin >> s;
    if (s == "white") return p[now].color = 0;
    else return p[now].color = 1;
}

int main() {
    int n;
    scanf("%d", &n);
    int l = 1, r = 1e9, mid;
    for (int i = 1; i <= n; i += 1) {
        mid = l + r >> 1;
        if (Check(mid, i)) r = mid;
        else l = mid;
    }
    int res = 0;
    for (int i = 1; i < n; i += 1)
        std:: swap(p[i], p[i + 1]);
    std::sort(p + 1, p + 1 + n);
    for (int i = 2; i <= n; i += 1)
        if (p[i].color != p[i - 1].color) {
            res = p[i - 1].position + 1;
            break;  
        }  
    if (res) printf("%d %d %d %d\n", XXX - 1, res - 1, XXX + 1, res), fflush(stdout);
    else printf("%d %d %d %d\n", XXX - 1, res, XXX + 1, res), fflush(stdout);

    return 0;
}
 

CF1064 E - Dwarves, Hats and Extrasensory Abilities