An easy problem(位运算)

As we known, data stored in the computers is in binary form.（数据以二进制形式存储于电脑之中。）The problem we discuss now is about the positive integers and its binary form.

Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of '1's in whose binary form is the same as that in the binary form of I.

For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 '1's. The minimum integer, which is greater than "1001110" and also contains 4 '1's, is "1010011", i.e. "83", so you should output "83".

#include<iostream>

using namespace std;

int count(int n)

{

	int res=0;

	while(n)

	{

		res++;

		n=n&(n-1);

	}

	return res;

}

int main()

{

	int n;

	while(cin>>n && n)

	{

		int cnt=count(n);

		while(n++)//自增

		{

			if(cnt==count(n))

			{

				cout<<n<<endl;

				break;

			}

		}

	}

	return 0;

}

巴特西

An easy problem(位运算)

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