ACM 第八届山东省赛 I Parity check SDUT 3901

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Parity check

Time Limit: 2000MS Memory Limit: 524288KB

Submit Statistic

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

Hint

Author

“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）

 找规律, 会发现%2   每三位 都是0 1 1   因此 看是 %3 余几即可 但又因为是10^1000  因此是个大数  ,大数处理 就行

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdio.h>
#include <string>

using namespace std;

int main()
{
    string s;
    int sum=0;
    int i;
    while(cin>>s)
    {
        int len=s.length();
        sum=0;
        for(i=0;i<len;i++)
        {
            sum+=(s[i]-'0')%3;
        }
        sum%=3;
        switch(sum)
        {
            case 1:printf("1\n");break;
            case 0:printf("0\n");break;
            case 2:printf("1\n");break;
        }

    }
    return 0;
}