UVA10931 Parity【进制】

We define the parity of an integer n as the sum of the bits in binary representation computed modulo two. As an example, the number 21 = 101012 has three 1s in its binary representation so it has parity 3(mod2), or 1.

  In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647.

Input

Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that should not be processed.

Output

For each integer I in the inputt you should print a line ‘The parity of B is P (mod 2).’, where B is the binary representation of I.

Sample Input

1

2

10

21

0

Sample Output

The parity of 1 is 1 (mod 2).

The parity of 10 is 1 (mod 2).

The parity of 1010 is 2 (mod 2).

The parity of 10101 is 3 (mod 2).

题链接：UVA10931 Parity

问题简述：（略）

问题分析：

  简单的10进制转2进制问题，再对2进制各位数字求和即可。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* UVA10931 Parity */

#include <bits/stdc++.h>

using namespace std;

const int N = 64;
char bits[N + 1];

int main()
{
    int n;
    while(~scanf("%d", &n) && n) {
        int i = 0, sum = 0;
        while(n) {
            sum += n & 1;
            bits[i++] = (n & 1) + '0';
            n >>= 1;
        }

        printf("The parity of ");
        while(i)
            putchar(bits[--i]);
        printf(" is %d (mod 2).\n", sum);
    }

    return 0;
}