Codeforces Round #260 (Div. 2)AB

http://codeforces.com/contest/456/problem/A

A. Laptops

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.

Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.

Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).

All ai are distinct. All bi are distinct.

Output

If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).

Sample test(s)

input

2
1 2
2 1

output

Happy Alex

------------------------------------------------------------------------------------------------

B. Fedya and Maths

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:

(1n + 2n + 3n + 4n) mod 5

for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

Input

The single line contains a single integer n (0 ≤ n ≤ 10105). The number doesn't contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Sample test(s)

input

4

output

4

input

124356983594583453458888889

output

0

Note

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

===============================================================================

#include <stdio.h>
#include <string.h>

int main()
{
    int n,m,i,j;
    int a,b,c,d;
    scanf("%d",&n);
    //scanf("%d%d",&a,&b);
    int sum=0;
    for(i=0;i<n;i++)
    {
        scanf("%d%d",&c,&d);
        if(c<d)
        sum++;
    }
    if(sum>0)
    printf("Happy Alex
");
    else
    printf("Poor Alex
");
}

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
char str[1000005];
int main()
{
    //
    int n,m,i,j;
    //memset(str,0,sizeof(str));
    while(scanf("%s",str)!=EOF)
    {
         int len=strlen(str);
         int data=0;
          if(len == 1)
          {
              if((str[len-1]-'0')%4==0)
              printf("4
");
              else printf("0
");
              continue;
          }
         data=(str[len-1]-'0')+(str[len-2]-'0')*10;
         if(data % 4 == 0)
         printf("4
");
         else
         printf("0
");
    }
    return 0;
}