COJ 1059 - Numeric Parity 位操作

非常好玩的一道题。能够熟悉下位操作实现和玩一玩bitset这个容器

Description

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 = 10101 has three 1s in its binary representation so it has parity 3 (mod 2), or 1.
In this problem you have to calculate the parity of an integer 1 <= I <= 2147483647 (2^31-1). Then, let start to work...

Input specification

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 specification

For each integer I in the input you should print one line in the form "The parity of B is P (mod 2)." where B is the binary representation of I.

Sample input

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).

使用bitset来实现。注意bitset的高低为存储顺序，是底位到高位。索引i右0到大的：

void NumericParity()

{

	int n = 0;

	bitset<32> bi;

	while (cin>>n && n)

	{

		bi = n;

		cout<<"The parity of ";

		bool flag = false;

		for (int i = bi.size() - 1; i >= 0 ; i--)

		{

			flag |= bi.test(i);

			if (flag) cout<<bi[i];

		}

		cout<<" is "<<bi.count()<<" (mod 2).\n";

	}

}

自家自制的位操作：

static bool biNum[32];

int intTobi(int n)

{

	int i = 0, c = 0;

	while (n)

	{

		c += n % 2;

		biNum[i++] = n % 2;

		n >>= 1;

	}

	return c;

}

void NumericParity2()

{

	int n = 0;

	while (cin>>n && n)

	{

		fill(biNum, biNum+32, false);

		cout<<"The parity of ";

		int c = intTobi(n);

		bool flag = false;

		for (int i = 31; i >= 0 ; i--)

		{

			flag |= biNum[i];

			if (flag) cout<<biNum[i];

		}

		cout<<" is "<<c<<" (mod 2).\n";

	}

}

巴特西