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

2015-10-22 16:05 429 查看
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,r

template<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;
}
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