Dylans loves numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 274 Accepted Submission(s): 181

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

题意很简单，判断一个数变成二进制之后，被0分开的1有多少个。

对一个数变为二进制之后，从右至左开始检验，前一个数是re，正在验的数是cur，如果满足re为0而cur为1，则分出一组，记录下来即可。

代码：

#include 
     
      #include 
      
       #include 
       
        using namespace std;long long result(long long s){    long long test=s;    int re=0;    long long sum=0;    while(test)    {        int cur = test&1;        if(re==0&&cur==1)        {            sum++;        }        re=cur;        test=test>>1;     }    return sum;}int main(){    int test;    cin>>test;    while(test--)    {        long long s;        cin>>s;        cout<
        
         <
        
       
      
     

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