第八届山东省赛题->Parity check

Problem Description

Fascinated with the computer games, Gabriel even forgets to study. Now she needs to finish her homework, and there is an easy problem:

f(n)=

She is required to calculate f(n) mod 2 for each given n. Can you help her?

Input

Multiple test cases. Each test case is an integer n(0≤n≤) in a single line.

Output

For each test case, output the answer of f(n)mod2.

Example Input

２

Example Output

１

看到f(n)的表达式是不是立刻想到斐波那契数列了？如果是，那就错了！关键是f(n) mod 2　这意味着输出只有０和１（就是让你找规律的！）　而且看到n的范围的么，１０的１０００次方，这是什么概念...　所以，输入就要变一变啦，用字符串接受输入的数，字符串怎么参与计算呢？这就需要观察我们找的规律啦，　０１１０１１０１１０...　看出来没...（n  mod 3)　所以，，现在关键是怎么把输入的字符串变n，既然是n mod 3，初中知识告诉我们...　一个数字能整除３，那数字的每一位上的数字之和也能整除３...　所以，问题解决啦，把字符串的每个字符变成数字然后相加，得到n,代码如下：

#include <stdio.h>#include <string.h>#define N 1010int main(){char s[N];int i,sum;while(~scanf("%s",s)){sum=0;for(i=0;i<strlen(s);i++)sum+=s[i]-'0';if(sum%3==2||sum%3==1)printf("1\n");elseprintf("0\n");}    return 0;}