poj2453 an easy problem (位运算)
2015-05-08 20:50
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An Easy Problem
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
Sample Output
Source
POJ Monthly,zby03
解析:http://blog.csdn.net/w57w57w57/article/details/6657547
代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7996 | Accepted: 4778 |
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
Source
POJ Monthly,zby03
解析:http://blog.csdn.net/w57w57w57/article/details/6657547
代码:
#include<cstdio> using namespace std; void init() { freopen("poj2453.in","r",stdin); freopen("poj2453.out","w",stdout); } void work() { int n,ans,t,x; while(scanf("%d",&n)&&n) { x=n&(-n),t=n+x; ans=t|((n^t)/x)>>2; printf("%d\n",ans); } } int main() { init(); work(); return 0; }
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