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
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
A line containing a number "0" terminates input, and this line need not be processed.
Output
Sample Input
1 2 3 4 78 0
Sample Output
2
4
5
8
我的一位同學竟用模擬!
無語中。。。
這是他的帖子:http://blog.csdn.net/chenwenxiaocom/archive/2011/05/16/6425414.aspx
- #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);
- }
- }
pascal翻譯應爲:
var n,x:longint; begin readln(n); while n<>0 do begin x:=n and -n; writeln(n+x+(n xor (n+x))div x div 4); readln(n); end; end.