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CF#260 B. Fedya and Maths

时间:2014-08-09 11:47:07      阅读:404      评论:0      收藏:0      [点我收藏+]

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

(1n?+?2n?+?3n?+?4nmod 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:

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不懂什么费马小定理。打表之后发现:n%4=0的时候输出4,
其余输出0,但n数据太大了.
我们知道:100以上的数可以分解成100*x+y者只需考虑y就可以了,
即只需考虑最后两位。这就好办了。读入字符串得到最后两位取模就可以了
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<cmath>
using namespace std;
const int maxn=1e6;
char s[maxn];
int main()
{
    while(~scanf("%s",s))
    {
        int len=strlen(s);
        int sum=0;
        sum+=s[len-1]-'0'+(s[len-2]-'0')*10;
        if(sum%4==0)
            cout<<4<<endl;
        else
            cout<<0<<endl;
    }
    return 0;
}



CF#260 B. Fedya and Maths,布布扣,bubuko.com

CF#260 B. Fedya and Maths

标签:style   http   color   os   io   数据   for   ar   

原文地址:http://blog.csdn.net/u013582254/article/details/38453601

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