CodeForces 416A Guess a number!
2015-10-02 16:05
513 查看
链接:http://codeforces.com/problemset/problem/416/A
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".
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.
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).
Output
Input
Output
直接模拟,附上AC代码:
#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef unsigned int li;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
int n, x;
char sign[5], answer;
int main()
{
ios::sync_with_stdio(false);
while (~scanf("%d", &n))
{
int maxy = 2000000000;
int miny = -2000000000;
while (n--)
{
scanf("%s%d %c", sign, &x, &answer);
if (!strcmp(">", sign))
{
if (answer == 'Y')
miny = max(miny, x+1);
else
maxy = min(maxy, x);
}
else if (!strcmp(">=", sign))
{
if (answer == 'Y')
miny = max(miny, x);
else
maxy = min(maxy, x-1);
}
else if (!strcmp("<", sign))
{
if (answer == 'Y')
maxy = min(maxy, x-1);
else
miny = max(miny, x);
}
else
{
if (answer == 'Y')
maxy = min(maxy, x);
else
miny = max(miny, x+1);
}
}
if (miny > maxy)
printf("Impossible\n");
else
printf("%d\n", miny);
}
return 0;
}
Guess a number!
time limit per test:1 secondmemory 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). Nextn 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 numbery 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)
Input4 >= 1 Y < 3 N <= -3 N > 55 N
Output
17
Input
2 > 100 Y < -100 Y
Output
Impossible
直接模拟,附上AC代码:
#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef unsigned int li;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
int n, x;
char sign[5], answer;
int main()
{
ios::sync_with_stdio(false);
while (~scanf("%d", &n))
{
int maxy = 2000000000;
int miny = -2000000000;
while (n--)
{
scanf("%s%d %c", sign, &x, &answer);
if (!strcmp(">", sign))
{
if (answer == 'Y')
miny = max(miny, x+1);
else
maxy = min(maxy, x);
}
else if (!strcmp(">=", sign))
{
if (answer == 'Y')
miny = max(miny, x);
else
maxy = min(maxy, x-1);
}
else if (!strcmp("<", sign))
{
if (answer == 'Y')
maxy = min(maxy, x-1);
else
miny = max(miny, x);
}
else
{
if (answer == 'Y')
maxy = min(maxy, x);
else
miny = max(miny, x+1);
}
}
if (miny > maxy)
printf("Impossible\n");
else
printf("%d\n", miny);
}
return 0;
}
相关文章推荐
- Transformation 能将 Windows XP/Server 2003 操作系统,完美地模拟成 Windows Vista 的软件
- 用javascript和css模拟select的脚本
- PHP模拟asp.net的StringBuilder类实现方法
- javascript用层模拟可移动的小窗口
- 自编jQuery插件实现模拟alert和confirm
- PHP模拟asp中response类实现方法
- javascript 模拟点击广告
- JQuery中模拟image的ajaxPrefilter与ajaxTransport处理
- php实现模拟post请求用法实例
- JavaScript实现MIPS乘法模拟的方法
- 模拟xcopy的函数
- JS模拟实现Select效果代码
- php模拟服务器实现autoindex效果的方法
- C# SendInput 模拟鼠标操作的实现方法
- PHP模拟登陆163邮箱发邮件及获取通讯录列表的方法
- JS模拟并美化的表单控件完整实例
- php模拟登陆的实现方法分析
- php模拟用户自动在qq空间发表文章的方法
- php 模拟 asp.net webFrom 按钮提交事件实例
- php模拟post提交数据的方法