Codeforces Round #241 (Div. 2) AGuess a number!

猜数字游戏逻辑解析

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本文介绍了一款名为“猜数字！”的游戏，详细解析了游戏规则及实现逻辑。玩家通过提问来猜测主持人设定的数字，而主持人则以是否的回答进行反馈。文章提供了具体的代码示例，演示如何根据一系列提问和答案找出可能的数字。

A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.

The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions:

Is it true that y is strictly larger than number x?
Is it true that y is strictly smaller than number x?
Is it true that y is larger than or equal to number x?
Is it true that y is smaller than or equal to number x?

On each question the host answers truthfully, "yes" or "no".

Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible".

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:

">" (for the first type queries),
"<" (for the second type queries),
">=" (for the third type queries),
"<=" (for the fourth type queries).

All values of x are integer and meet the inequation - 10⁹ ≤ x ≤ 10⁹. The answer is an English letter "Y" (for "yes") or "N" (for "no").

Consequtive elements in lines are separated by a single space.

Output

Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·10⁹ ≤ y ≤ 2·10⁹. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes).

Sample test(s)

Input

4
>= 1 Y
< 3 N
<= -3 N
> 55 N

Output

Input

2
> 100 Y
< -100 Y

Output

Impossible

注意数据范围：

设置两个极限，让下限取大的上限取小的。注意数据用%I64d!

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
#define LL __int64
#define inf 0x3f3f3f3f
using namespace std;
char now[10];
char s[3],c;
int main()
{
    LL  n,m,i,j,k,maxn,minn;
    while(~scanf("%I64d",&n)&&n)
    {
        maxn=2000000000;
        minn=-2000000000;
        getchar();
        while(n--)
        {
            scanf("%s %I64d",s,&m);
            getchar();
            scanf("%c",&c);
            if(!strcmp(">=",s))//3
            {
                if(c=='Y')
                {
                    minn=max(minn,m);
                }
                else
                {
                    maxn=min(maxn,m-1);
                }
            }
            else if(!strcmp("<=",s))//4
            {
                if(c=='N')
                {
                    minn=max(minn,m+1);
                }
                else
                {
                    maxn=min(maxn,m);
                }
            }
            else if(!strcmp(">",s))
            {
                if(c=='Y')
                {
                    minn=max(minn,m+1);
                }
                else
                {
                    maxn=min(maxn,m);
                }
            }
            else if(!strcmp("<",s))//2
            {
                if(c=='N')
                {
                    minn=max(minn,m);
                }
                else
                {
                    maxn=min(maxn,m-1);
                }
            }
        }
        if(maxn<minn)//只要是不满足这个条件就Impossible
            printf("Impossible\n");
       else
        {
            printf("%I64d\n",maxn);//否则就maxn<span id="transmark"></span>
        }
    }
    return 0;
}