zcmu 2009: Parity check

2009: Parity check

Time Limit: 2 Sec Memory Limit: 128 MB

Description

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

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≤10^1000) in a single line.
Output

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

2
Sample Output

1
HINT

Source

【分析】
咋一看，最先想到的是递归，但是后面多想了一下，首先n太大了，其次答案是要f(n)mod2的结果，既然要mod2，那么结果就只能是0或者1；这样想就发现其实是有规律所寻的（把f(0)~f(10)算出来，然后mod2看看，就能找到规律了）。

【代码】

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
 
char n[1010];
 
int main()
{
    while(scanf("%s",&n)!=EOF)
    {
        getchar();
        int ans=0,lenth=strlen(n);
        for(int i=0;i<lenth;i++)
        {
            int num=ans*10+(n[i]-'0');
            ans=num%3;
        }
        if(ans==1||ans==2)
            printf("1\n");
        else
            printf("0\n");
    }
    return 0;
}