山东省第八届acm大赛 I题 （SDUT 3901）

Parity check

Time Limit: 2000MS Memory Limit: 524288KB

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

规律题    能被3整除  为0    不能  为1         关键在于数据处理  10的1000次方      用到小学学到的知识       一个数能否被3整除  即这个数各位数字之和2能否被3整除

AC代码：

#include <stdio.h>#include <string.h>int main (){    char str[1005];    while (scanf ("%s",str)!=EOF){        long long sum=0;        int n=strlen(str);        for (int i=0;i<n;i++)            sum+=(str[i]-'0');        if (sum%3==0){            printf ("0\n");        }        else {            printf ("1\n");        }    }    return 0;}