POI 2453 你水的過嗎？僅以此獻給HYC小牛......

題目來源：  http://poj.org/problem?id=2453

 

 

 

 

An Easy Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6224		Accepted: 3655

Description

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".

Input

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.

Output

One integer per line, which is J.

Sample Input

1
2
3
4
78
0

Sample Output

2
4
5
8
83

你想用模擬嗎？我笑了。

你想用搜索嗎？我笑了。

你想用遺傳算法加深嗎？我笑了。

你還想構建log2(n)的圖來求嗎？我笑了。

讓你們看看”水“這個字的奧義吧（特別是某位小牛:-D）：

大聖代碼：

#include <stdio.h>
int main()
{
    int n,x;
    while(scanf("%d",&n),n)
    {
        x=n&-n;
        printf("%d/n",n+x+(n^n+x)/x/4);
    }
}

PAS如下：

program e2453;
var     i,j,n:longint;
begin   readln(n);
        while n<>0 do
        begin
         i:=n and (-n);
         writeln((n xor (n+i))div(4*i)+n+i);
         read(n);
        end;
end.                             

我的比較短而且有效......