时间:2021-07-01 10:21:17 帮助过:31人阅读
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
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 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 <cstdio>
- #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();
- }</cstdio>
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