2012 Multi-University Training Contest 5:Xiao Ming's Hope

最新推荐文章于 2018-08-14 15:24:12 发布

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最新推荐文章于 2018-08-14 15:24:12 发布

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小明在2011年11月11日这天，面对着数学课本上关于组合数的问题，尝试计算组合数中奇数的数量。在遇到困难时，他请求程序员的帮助来解决这个看似简单的数学问题。本文将揭示如何通过计算2^n * 2^(二进制中1的个数)来找出给定n下组合数中奇数的个数。

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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).

SampleInput

1
2
11

SampleOutput

2
2
8

//今天和队友一起做多校联合5，共切了5题，和电大神牛队9题比起来我们弱爆了！

//本题其实就是求C(n,0)....C(n,n)中奇数的个数。由Lucas定理推演可以知道其实就是求（表达式的和2^n）2^（n中2进制1的个数）。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int m=0;
        while(n)
        {
            if(n%2) m++;
            n/=2;
        }
        printf("%I64d\n",(__int64)pow(2.0,m));
    }
    return 0;
}