Fedya and Maths

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:

 

这道题上来直接打表找规律

发现能整除4的 n 算出来是4，其他的n为0。要注意输入很大，要用字符串。

这里处理就出多说了，假设一个数字n前面n-2个数为 x 后面两个数字为 y ，则这个数 n=x*100+y

前面必然能除4，判断后面两位能否整除四即可

 1 #include<cstdio>

 2 #include<cmath>

 3 #include<cstring>

 4 #include<stdlib.h>

 5 using namespace std;

 6 int main()

 7 {

 8     char str[100005],a,b;

 9     int ans;

10     scanf("%s",str);

11     int len=strlen(str);

12     if(len>=2)

13     {

14         a=str[len-1];

15         b=str[len-2];

16         ans=(b-'0')*10+a-'0';

17     }

18     else

19         ans=str[len-1]-'0';

20 

21     if(ans%4==0)

22         printf("4\n");

23     else

24         printf("0\n");

25     return 0;

26 }

View Code