HDU 4349 Xiao Ming ' s Hope,lucas theorem

Xiao Ming ' s Hope Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1793 Accepted Submission (s): 1189


Problem Description Xiao Ming likes counting numbers very much, especially he is fond of counting odd numbers. Maybe he thinks it is the best-of-the-show he is alone without a girl friend. The day 2011.11.11 comes. Seeing classmates walking with their girl friends, he coundn ' t help running to he classroom, and then opened his maths Book preparing to count odd numbers. He looked at his book, then he found a question "C (n,0) +c (n,1) +c (n,2) +...+c (n,n) =?". Of course, Xiao Ming knew the answer, but he didn ' t care on that, what's he wanted to know is, how many odd numbers There were? Then he began to count odd numbers. When n was equal to 1, C (1,0) =c (=1), there was 2 odd numbers. When n was equal to 2, C (2,0) =c (2,2) =1, there was 2 odd numbers ... Suddenly, he found a girl was watching him counting odd numbers. In order to show his gifts on maths, he wrote several big numbers what n would is equal to, but he found it was impossible To finished him tasks, then he sent a piECE of information to you, and wanted your a excellent programmer to help him, he really didn ' t want to let she down. Can you help him?  
Input each line contains a integer n (1<=n<=10 8)
Output A single line with the number of odd numbers of C (n,0), C (n,1), C (n,2) ... C (N,n).
Sample Input

1 2 11
Sample Output

2 2 8
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#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
int main () {
    long long n;
    while (cin>>n) {
        int res = 0;
        while (n) {
            if (n&1) res++;
            n/=2;
        }
        A long long ans = 1;
        for (int i = 0;i < res;i++)
            ans*=2;
        cout<<ans<<endl;
    }
    return 0;
}