1、http://codeforces.com/problemset/problem/416/A
2、题目：
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".
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.
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).
4 >= 1 Y < 3 N <= -3 N > 55 N
17
2 > 100 Y < -100 Y
Impossible3、AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 2*1000000000
int main()
{int n,x;char str[5];char ch;int a;int b;while(scanf("%d",&n)!=EOF){a=-INF;b=INF;int flaga=0,flagb=0;//0表示开区间while(n--){scanf("%s%d",str,&x);getchar();scanf("%c",&ch);int l=strlen(str);//printf("%s %d %c\n",str,x,ch);if(l==1){if(str[0]=='>'){if(ch=='Y' && x>=a){a=x;flaga=0;}else if(ch=='N' && x<b){b=x;flagb=1;}}else if(str[0]=='<'){if(ch=='Y' && x<=b){b=x;flagb=0;}else if(ch=='N' && x>a){a=x;flaga=1;}}}else if(l==2){if(str[0]=='>'){if(ch=='Y' && x>a){//printf("a=%d\n",a);a=x;flaga=1;}else if(ch=='N' && x<=b){b=x;flagb=0;}}else if(str[0]=='<'){if(ch=='Y' && x<b){b=x;flagb=1;}else if(ch=='N' && x>=a){a=x;flaga=0;}}}}if(flaga==1 && flagb==1 && b>=a)printf("%d\n",a);else if(flaga==1 && flagb==0 && b>a){printf("%d\n",a);}else if(flaga==0 && flagb==1 && a<b){printf("%d\n",b);}else if(flaga==0 && flagb==0){if(a!=-INF && b>=a+2){printf("%d\n",a+1);}else if(b!=INF && b>=a+2){printf("%d\n",b-1);}elseprintf("Impossible\n");}else{printf("Impossible\n");}}return 0;
}
/*
1
< 1000000000 Y
*/
/*
4
>= 1 Y
< 3 N
<= -3 N
> 55 N
2
> 100 Y
< -100 Y
1
>= 3 Y
1
>= 3 N
1
> 3 N
1
> 3 Y
1
< 3 Y
1
< 3 N
1
<= 3 N
1
<= 3 Y
*/