HDU 5272 Dylans loves numbers

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5272

题面：

Dylans loves numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 241    Accepted Submission(s): 160

Problem Description

Who is Dylans?You can find his ID in UOJ and Codeforces.
His another ID is s1451900 in BestCoder.

And now today's problems are all about him.

Dylans is given a number  N .
He wants to find out how many groups of "1" in its Binary representation.

If there are some "0"（at least one）that are between two "1",
then we call these two "1" are not in a group,otherwise they are in a group.

 

Input

In the first line there is a number  T .

T  is the test number.

In the next  T  lines there is a number  N .

0≤N≤1018,T≤1000

 

Output

For each test case,output an answer.

 

Sample Input

   
   
   
   

    
    
    
    1
5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2
   
   
   
   

 

Source

BestCoder Round #45

 

解题：

就是求将一个数转成二进制后，其中有几块“1”。因为首位肯定为1，所以只要往后多加一个“0”。然后数连续“10”的个数即可。

代码：

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
	int t,x,cnt;
	unsigned long long n;
	string s;
	cin>>t;
	while(t--)
	{
		cin>>n;
		cnt=0;
		s="";
		while(n)
		{
		  x=n&1;
		  if(x)	s+='1';
		  else s+='0';
		  n>>=1;
		}
		reverse(s.begin(),s.end());
		s+='0';
		int len=s.length()-1;
		for(int i=0;i<len;i++)
		{
			if(s[i]=='1'&&s[i+1]=='0')
			cnt++;
		}
		cout<<cnt<<endl;
	}
	return 0;
}