Parity 进制转换

最新推荐文章于 2024-04-10 20:16:01 发布

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最新推荐文章于 2024-04-10 20:16:01 发布

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本文链接：https://blog.youkuaiyun.com/Wonder__/article/details/79744548

该问题要求计算1到2^31-1之间的整数在二进制表示下的奇偶性，即计算二进制表示中1的个数对2取模的结果。给出的C++代码通过递归函数实现进制转换并统计1的个数，然后输出结果。

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We define the parity of an integer n as the sum of the bits in binary representation computed modulotwo.

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

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).

#include<bits/stdc++.h>
using namespace std;

int ans;

void zh(int x)///递归转化进制
{
if(x == 0) return;
int k = x%2;
if(k == 1) ans++;
x = x/2;
zh(x);
printf("%d",k);
}

int main()
{
int n;
while(~scanf("%d",&n)&&(n!=0))
{
ans = 0;
printf("The parity of ");
zh(n);
printf(" is %d (mod 2).\n",ans);
}
}