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; }