CodeForces - 416A Guess a number

A. Guess a number!
time limit per test1 second
memory limit per test256 megabytes

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 - 10^9 ^≤ 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).

Examples
input
4
>= 1 Y
< 3 N
<= -3 N
> 55 N
output
17
input
2
> 100 Y
< -100 Y
output
Impossible

问题链接：CodeForces - 416A Guess a number!
问题简述：（略）
问题分析：（略）

AC的C语言程序如下：

/* CodeForces - 416A Guess a number! */#include <stdio.h>#define Y 2000000000int main()
{int n, x, miny = -Y, maxy = Y;char sign[3], ans[2];scanf("%d", &n);while (n--) {scanf("%s%d%s", sign, &x, ans);if (ans[0] == 'Y') {if (sign[0] =='>' && sign[1] == '=') {if (miny < x) miny = x;} else if (sign[0] == '>') {if (miny <= x) miny = x + 1;} else if (sign[0] == '<' && sign[1] == '=') {if (maxy > x) maxy = x;} else if (sign[0] == '<') {if (maxy >= x) maxy = x - 1;}} else if (ans[0] == 'N') {if (sign[0] =='<' && sign[1] == '=') {if (miny <= x) miny = x + 1;} else if (sign[0] == '<') {if (miny < x) miny = x;} else if (sign[0] == '>' && sign[1] == '=') {if (maxy >= x) maxy = x - 1;} else if (sign[0] == '>') {if (maxy > x) maxy = x;}}}if (miny <= maxy) printf("%d\n", miny);else printf("Impossible\n");return 0;
}

AC的C++语言程序如下：

/* CodeForces - 416A Guess a number! */#include <bits/stdc++.h>using namespace std;const int Y = 2000000000;
int main()
{ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);int n, x, miny = -Y, maxy = Y;string sign;char ans;cin >> n;while (n--) {cin >> sign >> x >> ans;if (ans == 'N') {if (sign == ">") sign = "<=";else if (sign == "<") sign = ">=";else if (sign == ">=") sign ="<";else if (sign == "<=") sign = ">";}if (sign == ">" && miny <= x) miny = x + 1;else if (sign == "<" && maxy >= x) maxy = x - 1;else if (sign == ">=" && miny < x) miny = x;else if (sign == "<=" && maxy > x) maxy = x;}if (miny <= maxy)cout << miny << endl;elsecout << "Impossible" << endl;return 0;
}

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