codeforcesRound#241(div2)A解题报告

A. Guess a number! time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A TV show called Guess a number! is gathering popularity. The whole Berland, the old and the young, are watchin

A. Guess a number!

time limit per test

memory limit per test

input

output

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:

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:

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

17

input

2
> 100 Y
< -100 Y

output

Impossible

题目大意:

猜数字,给出n个区间询问,然后同时给出该区间是否正确,然后求出任何符合上述区间要求的数字.

解法:

先将答案范围规定为: [l,r] ;其中 l = -2*10^9 r = 2*10^9

每次读入,更新一次上届和下届,l和r;

读入结束后,

l > r Impossible

l <= r 随意输出一个区间内的数字即可

代码:

#include 
#define INF 2000000000

int n;

int max(int a, int b) {
	if (a > b) return(a);
	return(b);
}

int min(int a, int b) {
	if (a < b) return(a);
	return(b);
}

void init() {
	scanf("%d\n", &n);
}

void solve() {
	int l = -INF, r = INF, x;
	char ch1, ch2, ch3;

	for (int i = 1; i <= n; i++) {
	ch1 = getchar();
	ch2 = getchar();

	scanf("%d ", &x);
	ch3 = getchar();
	getchar();

	if (ch3 == 'Y') {
	if (ch1 == '>') {
	if (ch2 == '=')
	l = max(x, l);
	else
	l = max(x+1, l);
	}

	if (ch1 == '<') {
	if (ch2 == '=')
	r = min(r, x);
	else
	r = min(r, x-1);
	}
	}

	if (ch3 == 'N') {
	if (ch1 == '>') {
	if (ch2 == '=')
	r = min(r, x-1);
	else
	r = min(r, x);
	}

	if (ch1 == '<') {
	if (ch2 == '=')
	l = max(l, x+1);
	else
	l = max(l, x);
	}
	}

	}

	if (l <= r)
	printf("%d", l);
	else
	printf("Impossible");
}

int main() {
	init();
	solve();
}

.

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