• Codeforces Round #534 (Div. 2) D. Game with modulo 交互题


    先二分一个区间,再在区间里面二分即可;

    可以仔细想想,想明白很有意思的;

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 200005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    //const long long int mod = 1e9 + 7;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-4
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    //const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    inline ll rd() {
    	ll x = 0;
    	char c = getchar();
    	bool f = false;
    	while (!isdigit(c)) {
    		if (c == '-') f = true;
    		c = getchar();
    	}
    	while (isdigit(c)) {
    		x = (x << 1) + (x << 3) + (c ^ 48);
    		c = getchar();
    	}
    	return f ? -x : x;
    }
    
    ll gcd(ll a, ll b) {
    	return b == 0 ? a : gcd(b, a%b);
    }
    int sqr(int x) { return x * x; }
    
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
    	if (!b) {
    		x = 1; y = 0; return a;
    	}
    	ans = exgcd(b, a%b, x, y);
    	ll t = x; x = y; y = t - a / b * y;
    	return ans;
    }
    */
    
    string s;
    
    int main() {
    	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    	while (1) {
    		cin >> s;
    		if (s == "end") {
    			return 0;
    		}
    		if (s == "start") {
    			ll a = 1;
    			for (; a < (1 << 30); a <<= 1) {
    				cout << "?" << ' ' << a / 2 << ' ' << a << endl;
    				fflush(stdout);
    				cin >> s;
    				if (s == "x") {
    					break;
    				}
    			}
    			// 区间 a/2~ a
    			ll l = a / 2, r = a;
    			while (l + 1 < r) {
    				ll mid = (l + r) >> 1;
    				cout << "?" << ' ' << mid << ' ' << l << endl;
    				fflush(stdout);
    				cin >> s;
    				if (s == "x") {
    					l = mid;
    				}
    				else r = mid;
    			}
    			cout << "!" << ' ' << r << endl;
    			fflush(stdout);
    		}
    	}
    }
    
    EPFL - Fighting
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10307340.html
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