hdu-An Easy Problem

http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1002&ojid=1&cid=5570&hide=0

[align=left]Problem Description[/align]
As we known, data stored in the computers is in binary form. 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".

[align=left]Input[/align]
One integer per line, which is I (1 <= I <= 1000000).

A line containing a number "0" terminates input, and this line need not be processed.

[align=left]Output[/align]
One integer per line, which is J.

[align=left]Sample Input[/align]

1
2
3
4
78
0

[align=left]Sample Output[/align]

2
4
5
8
83

分析： 队列做法
#include <iostream>
#include <cmath>
#include <queue>
#include <cstdio>
using namespace std;
int bin[100];
int change(int num)
{
int k=0,l;
queue<int>z;

if(num==0)
bin[0]=0;
else
{
while(num>0)
{
z.push(num%2);
num=num/2;
}
while(!z.empty())
{
bin[k]=z.front();
k++;
z.pop();
}

l=k;

}
return l;
}
int main()
{
int a,b,count,i,k,j,p;
while(cin>>a)
{
if(a==0)break;
count=0;
k=change(a);
bin[k]=0;
for(i=0; i<k; i++)
{
if(bin[i]==1&&bin[i+1]==0)
{

bin[i]=0;
bin[i+1]=1;
break;
}
else
{
if(bin[i]==1)
{
count++;
bin[i]=0;
}
}
}
for(i=0; i<count; i++)
bin[i]=1;
count=0;
for(i=k; i>=0; i--)
{
count+=bin[i]*pow(2.0,i);
}
cout<<count<<endl;
}
return 0;
}

位运算做法：[code]#include <iostream>
using namespace std;
const int MAX_BIT=21;
int CountOne(int input)
{
int cnt=0;
for (int i=0;i<MAX_BIT;i++)
{
if(input>>i&1)//“>>”右移一位。本句话的意思是十进制数换成二进制数然后右移一位，判断是不是1，如果是cnt+1
cnt++;
}
return cnt;
}
int main()
{
int input;
while (cin>>input&&input!=0)
{
for (int ii=input+1;;ii++)
{
if(CountOne(ii)==CountOne(input))
{
cout<<ii<<endl;
break;
}
}
}
return 1;
}

[/code]