hdu 4349 Xiao Ming's Hope

Xiao Ming's Hope

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1531    Accepted Submission(s): 1028

Problem Description

Xiao Ming likes counting numbers very much, especially he is fond of counting odd numbers. Maybe he thinks it is the best way to 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 into his 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 about that , What he wanted to know was that how many odd numbers there were? Then he began to count odd numbers. When n is equal to 1, C _(1,0)=C _(1,1)=1, there are 2 odd numbers. When n is equal to 2, C _(2,0)=C _(2,2)=1, there are 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 be equal to, but he found it was impossible to finished his tasks, then he sent a piece of information to you, and wanted you a excellent programmer to help him, he really didn't want to let her 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
  

 

Author

HIT

 

Source

2012 Multi-University Training Contest 5

题目分析：
要求这些组合数中质数的个数，那么也就是C(n,m)%2 == 1 的个数，首先C(n,m)%2 等于C(n%2,m%2)*(C(n/2,m/2)%2,而只有C(0,1) = 0 ,所以当n的某一二进制位为0时,m的必为0，那么考虑为n的二进制位为1的时候，m的二进制既可以为1也可以为0就能保证形成的数是奇数，所以最终结果是2^(n的二进制位中1的个数）
代码是幼儿园小孩都能1A的

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

int n;

int main ( )
{
    while ( ~scanf ( "%d" , &n ) )
    {
        int cnt = 0;
        while ( n )
        {
            if ( n&1 ) cnt++;
            n >>= 1;
        }
        printf ( "%d\n" , 1<<cnt );
    }
}