Gym 100792C Colder-Hotter (三分)

题意:系统有一个点对,让你去猜,每次你猜一个,如果这个数和系统里的那个点距离比上一个你猜的近,那么返回1,否则返回0,第一次猜一定返回0,在不超过500次的情况下,猜出正确答案。

析:是一个简单的三分,横纵坐标可以分开来考虑,每次两次三分,然后看那个点更偏向哪边即可,注意这个题,有一个坑,那就是你输出的点不能超过1e9,即使是在运算过程中也不行,也就是说上界必须是  1e9,其他的都不对,我就写了1e9+1,WA到死。。。。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
#define pii pair<int, int>
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 18260 + 100;
const LL mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r > 0 && r <= n && c > 0 && c <= m;
}


int main(){
  int ansx = 0, ansy = 1e9;
  int cnt = 0, x1, x2;
  int idx = 0;
  while(ansx < ansy){
    int m1 = ansx + (ansy - ansx) / 3;
    int m2 = ansy - (ansy - ansx) / 3;
    cout << 0 << " " << m1 << endl << endl;
    cout.flush();
    cin >> x1;

    cout << 0 << " " << m2 << endl << endl;
    cout.flush();
    cin >> x2;
    if(x2 == 1)  ansx = m1 + 1;
    else ansy = m2 - 1;
    cnt += 2;
    if(cnt > 500)  assert(0);
  }
  int ansxx = ansx;
  ansx = 0, ansy = 1e9;
  idx = 0;
  while(ansx < ansy){
    int m1 = ansx + (ansy - ansx) / 3;
    int m2 = ansy - (ansy - ansx) / 3;
    cout << m1 << " " << ansxx << endl << endl;
    cout.flush();
    cin >> x1;

    cout << m2 << " " << ansxx << endl << endl;
    cout.flush();
    cin >> x2;
    if(x2 == 1)  ansx = m1 + 1;
    else ansy = m2 - 1;
    cnt += 2;
    if(cnt > 500)  assert(0);
  }
  cout << "A " << ansx << " " << ansxx << endl;
  return 0;
}

  

原文地址:https://www.cnblogs.com/dwtfukgv/p/7572784.html