2015-2016 ACM-ICPC, NEERC, Moscow Subregional Contest C. Colder-Hotter

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交互题

首先三分x坐标，然后因为三分不准确，所以在附近震荡求精确值。
其次同样的方法求出y坐标。

注意，询问次数的上限是500。
每次询问的时候， $(x,y)$ 的两个坐标必须在 $[0,1E9]$ 之间。

//      whn6325689
//      Mr.Phoebe
//      http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
#define speed std::ios::sync_with_stdio(false);typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))
#define getidx(l,r) (l+r|l!=r)
#define ls getidx(l,mid)
#define rs getidx(mid+1,r)
#define lson l,mid
#define rson mid+1,rtemplate<class T>
inline bool read(T &n)
{T x = 0, tmp = 1;char c = getchar();while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();if(c == EOF) return false;if(c == '-') c = getchar(), tmp = -1;while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();n = x*tmp;return true;
}
template <class T>
inline void write(T n)
{if(n < 0){putchar('-');n = -n;}int len = 0,data[20];while(n){data[len++] = n%10;n /= 10;}if(!len) data[len++] = 0;while(len--) putchar(data[len]+48);
}
//-----------------------------------
int ansx, ansy;int main()
{double xl = 0, xr = 1e9, yl = 0, yr = 1e9;for (int t = 1; t <= 100; t++){double mid1 = (xl+xr)/2, mid2 = (mid1+xr)/2;int flag1, flag2;printf("%.0lf 0\n", mid1);fflush(stdout);scanf("%d", &flag1);printf("%.0lf 0\n", mid2);fflush(stdout);scanf("%d", &flag2);if (flag2){xl = mid1;ansx = xl;}else{xr = mid2;ansx = xr;}}int tmp = ansx;for (int i = max(0, tmp-10); i <= min(1000000000, tmp+10); i++){int flag;printf("%d 0\n", i);fflush(stdout);scanf("%d", &flag);if (flag) ansx = i;}for (int t = 1; t <= 100; t++){double mid1 = (yl+yr)/2, mid2 = (mid1+yr)/2;int flag1, flag2;printf("%d %.0lf\n", ansx, mid1);fflush(stdout);scanf("%d", &flag1);printf("%d %.0lf\n", ansx, mid2);fflush(stdout);scanf("%d", &flag2);if (flag2){yl = mid1;ansy = yl;}else{yr = mid2;ansy = yr;}}tmp = ansy;for (int i = max(0, tmp-10); i <= min(1000000000, tmp+10); i++){int flag;printf("%d %d\n", ansx, i);fflush(stdout);scanf("%d", &flag);if (flag) ansy = i;}printf("A %d %d\n", ansx, ansy);return 0;
}