題意
(n)次操作,每次你給出一個點的座標,系統會返回該點的顏色(黑 / 白),程式最後輸出一條直線把所有黑點和白點分隔開
Sol
一個很直觀的想法:首先詢問((dx, 0)),然後每次詢問二分中點,根據與第一次詢問得到的字串的關係不斷調整二分範圍
但是這樣會被卡,我修改了兩個地方才過。
-
二分調整邊界的時候直接設(l = mid)或(r = mid),因為我們最後得到的不是一個精確解,所以這樣寫是可以的
-
最後輸出直線的時候加一個偏移量,也就是輸出一條斜線
具體看程式碼
/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + `0`;}
//#define OS *O++ = ` `;
//#define fout fwrite(obuf, O-obuf, 1 , stdout);
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 2005, INF = 1e9 + 10, mod = 1e9 + 7;
const int D[] = { -1, 1};
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
while(c >= `0` && c <= `9`) x = x * 10 + c - `0`, c = getchar();
return x * f;
}
int N, Dx = 23333;
string s, pre;
main() {
N = read();
int l = 0, r = 1e9;
printf("%d 0
", Dx);
fflush(stdout);
cin >> pre;
int ans = 0;
for(int i = 2; i <= N; i++) {
int mid = l + r >> 1;
printf("%d %d
", Dx, mid);
fflush(stdout);
cin >> s;
if(s != pre) r = mid;
else l = mid, ans = mid;
}
printf("%d %d %d %d", Dx - 3, ans, Dx + 3, ans + 1);
return 0;
}
/*
5
black
black
white
white
black
*/