题目链接：?pid=5272
题面：
Dylans loves numbersTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 241Accepted 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.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.is the test number.In the nextlines there is a number.
Output
For each test case,output an answer.
Sample Input
15
Sample Output
2
Source
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;}
，孤单寂寞与被遗弃感是最可怕的贫穷
