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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

題意：C（n,m)表示組合數n個取m的方案數；

　　　求C(n,0),C(n,1),C(n,2),......C(n,n)有多少個奇數；

思路：打表找規律，

　　　證明：盧卡斯　　　

　　　　C(n,m)%p=C(n/p,m/p)*C(n%p,m%p)%p;

　　　令p等於2，C(n%p,m%p) 為C(1,1),C(1,0),C(0,0),C(0,1) ,

　　　枚舉n的二進制表示每一位數，當前位為1是可以有兩種方案；

#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;
}

hdu 4349 Xiao Ming's Hope 規律