Codeforces Round #534 (Div. 2)D. Game with modulo-1104-D（交互+二分+构造）

This is an interactive problem.

Vasya and Petya are going to play the following game: Petya has some positive integer number

"x", if
"y", if

We define

Vasya should guess the number

It's guaranteed that Petya has a number, that satisfies the inequality

Help Vasya playing this game and write a program, that will guess the number

Interaction

Your program should play several games.

Before the start of any game your program should read the string:

"start" (without quotes) — the start of the new game.
"mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
"end" (without quotes) — all games finished. Your program should terminate after reading this string.

After reading the string "start" (without quotes) the new game starts.

At the beginning, your program should ask several questions about pairs of non-negative integer numbers

"x" (without quotes), if
"y" (without quotes), if
"e" (without quotes) — you asked more than

After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number

We recall that your program can't ask more than

If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.

Don't forget to flush the output after printing questions and answers.

To flush the output, you can use:

fflush(stdout) in C++.
System.out.flush() in Java.
stdout.flush() in Python.
flush(output) in Pascal.
See the documentation for other languages.

It's guaranteed that you should play at least

Hacks:

In hacks, you can use only one game. To hack a solution with Petya's number

Example

input

Copy

start
x
x
start
x
x
y
start
x
x
y
y
end

output

Copy

Note

In the first test, you should play

In the first game, Petya will answer "x" (without quotes) to any question, because

In the second game, if you will ask pair

极限数据情况下正好60次，所以算法是合理的，具体细节比较多，可以参考我的AC代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
char s[maxn];
char re[1000];
char ask(ll j,ll k)
{
    k=min(k,2000000000ll);
    printf("? %lld %lld\n",j,k);
    fflush(stdout);
    scanf("%s",re);
    return re[0];
}
int main()
{
    int o=-1;
    while(1)
    {

        scanf("%s",s);
        if(s[0]=='e'||s[0]=='m')
        {
            break;
        }else
        {
            char w=ask(0ll,1ll);
            if(w=='x')
            {
                printf("! 1\n");
                fflush(stdout);
            }else
            {
                ll j=1;ll k=2;
                int flag=0;
                repd(i,1,30)
                {
                    w=ask(j,k);
                    if(w=='x')
                    {
                        flag=1;
                        break;
                    }else
                    {
                        j=k;
                        k*=2;
                    }
                }
                if(flag)
                {
                    ll mid=(j+k)>>1;
                    repd(i,1,29)
                    {
                        mid=(j+k)>>1;
                        w=ask(mid,k);
                        if(w=='x')
                        {
                            j=mid;
                        }else
                        {

                            k=mid;

                        }
                    }
                    printf("! %d\n",max(j,k));
                    fflush(stdout);
                }
            }

        }
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

posted @ 2019-01-23 14:00 茄子Min 阅读(556) 评论(0) 编辑收藏举报

刷新页面返回顶部

茄子Min

起舞弄清影，何似在人间。

Codeforces Round #534 (Div. 2)D. Game with modulo-1104-D（交互+二分+构造）

公告