HDU 5272 Dylans loves numbers Dylans loves numbers

81 阅读 0 评论 54 点赞

我是靠谱客的博主自信金针菇，最近开发中收集的这篇文章主要介绍HDU 5272 Dylans loves numbers Dylans loves numbers，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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

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;
}