山东省第八届ACM省赛 I 题（Parity check）

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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.

//这道题一看数据范围就是到，不能用常规的递归来求解，找规律啊 啊啊啊啊啊

山东省第八届ACM省赛 I 题（Parity check）

#include <stdio.h>
#include<iostream>
#include <string.h>
#include<stdlib.h>
#include<string>
#include<cmath>
#include<algorithm>
#include<stdlib.h>
#define M 10005
using namespace std;
/*shandong ACM 8 I*/
//先按照普通方法找规律
/*int function(int n){
	if(n==0){
		return 0;
	}
	if(n==1){
		return 1;
	}
	return function(n-1)+function(n-2);
} 
 
int main(){
	
	int n;
	while(1){
		
	scanf("%d",&n);
	printf("%d\n",function(n)%2);
}
	return 0;
	 

}*/
//通过下面程序发现  if(n%3==0)  return 0  else return 1 
//接下来解决大数输入的问题  10的1000次方 
 
 
int main(){
	char c[1010];
	while(scanf("%s",c)!=EOF){
		int len=strlen(c);
		int sum=0;
		//判断每一位是否能被三{整除，若有一位不能被三整除，则    return 1
		for(int i=0;i<len;i++) {
			sum=(sum*10+(c[i]-'0'))%3; //实际上是从第一位对每一位取余数，当有余数时在对下一位取余数进位； 
		}
		sum==0?printf("0\n"):printf("1\n");
		return 0;
		
	} 
	
	
	

	return 0;
	 

}