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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…
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.
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.
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
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";
}
}
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