COJ 1059 – Numeric Parity 位操作「建议收藏」

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

Input specification

Output specification

Sample input

1210210

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";
	}
}