Parity check

Parity check

Time Limit: 2000 ms Memory Limit: 524288 KiB

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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.

Sample Input

2

Sample Output

1

1.数据太大，不可以用整形的来解决的时候，可以考虑字符串，这个已经说了好多次了。

2.还有就是，当循环次数太多，递归思路太多的时候，可以考虑，先把一些结果打印出来，观察观察是否有什么规律。

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