An Easy Problem --- 贪心

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

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".

输入

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.

输出

One integer per line, which is J.

样例输入

1
2
3
4
78
0

样例输出

2
4
5
8
83

题意：

随意输入一个字，比如78他的二进制是：1001110，要求输出一个数比这个78大的并且二进制数中1的个数是相同的最小的一个。

解题分析：

这用的是贪心法，每次求一的个数，然后每次在原来的数上加一，看看哪个数复合。

代码：

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int p, x, num1, num2;
while(1)
{
num1 = num2 = 0;
scanf("%d", &x);
if(!x)
break;
p = x;
while(p > 0)
{
if(p%2 != 0)
num1++;
p >>= 1;
}


while(1)
{
x++;
p = x;
num2 = 0;
while(p > 0)
{
if(p % 2 != 0)
num2++;
p >>= 1;
}
if(num2 == num1)
{
printf("%d\n", x);
break;
}
}
}
return 0;
}