hdu 4349 Xiao Ming's Hope 规律

Xiao Ming's Hope

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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<=10⁸)

 

 

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

题意：C（n,m)表示组合数n个取m的方案数；

　　　求C(n,0),C(n,1),C(n,2),......C(n,n)有多少个奇数；

思路：打表找规律，

　　　证明：卢卡斯　　　

　　　　C(n,m)%p=C(n/p,m/p)*C(n%p,m%p)%p;

　　　令p等于2，C(n%p,m%p) 为C(1,1),C(1,0),C(0,0),C(0,1) ,

　　　枚举n的二进制表示每一位数，当前位为1是可以有两种方案；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-8
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e2+10,M=2e6+10,inf=1e9+10;
const LL INF=1e18+10,mod=1e9+7;


int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int ans=0;
        while(n)
        {
            if(n&1)ans++;
            n/=2;
        }
        cout<<(1<<ans)<<endl;
    }
    return 0;
}