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codeforcesRound#241(div2)A解题报告

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:

  • 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

  1. 4
  2. >= 1 Y
  3. < 3 N
  4. <= -3 N
  5. > 55 N

output

  1. 17

input

  1. 2
  2. > 100 Y
  3. < -100 Y

output

  1. 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:

  1. #include <cstdio>
  2. #define INF 2000000000
  3. int n;
  4. int max(int a, int b) {
  5. if (a > b) return(a);
  6. return(b);
  7. }
  8. int min(int a, int b) {
  9. if (a < b) return(a);
  10. return(b);
  11. }
  12. void init() {
  13. scanf("%d\n", &n);
  14. }
  15. void solve() {
  16. int l = -INF, r = INF, x;
  17. char ch1, ch2, ch3;
  18. for (int i = 1; i <= n; i++) {
  19. ch1 = getchar();
  20. ch2 = getchar();
  21. scanf("%d ", &x);
  22. ch3 = getchar();
  23. getchar();
  24. if (ch3 == 'Y') {
  25. if (ch1 == '>') {
  26. if (ch2 == '=')
  27. l = max(x, l);
  28. else
  29. l = max(x+1, l);
  30. }
  31. if (ch1 == '<') {
  32. if (ch2 == '=')
  33. r = min(r, x);
  34. else
  35. r = min(r, x-1);
  36. }
  37. }
  38. if (ch3 == 'N') {
  39. if (ch1 == '>') {
  40. if (ch2 == '=')
  41. r = min(r, x-1);
  42. else
  43. r = min(r, x);
  44. }
  45. if (ch1 == '<') {
  46. if (ch2 == '=')
  47. l = max(l, x+1);
  48. else
  49. l = max(l, x);
  50. }
  51. }
  52. }
  53. if (l <= r)
  54. printf("%d", l);
  55. else
  56. printf("Impossible");
  57. }
  58. int main() {
  59. init();
  60. solve();
  61. }</cstdio>



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