hdu 4349 Xiao Ming's Hope，lucas定理

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#hdu 4349 #Xiao Mings Hope #lucas定理

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本题描述了小明面对一个数学挑战的过程：计算从C(n,0)到C(n,n)中奇数的数量。输入为一个整数n，范围在1到10^8之间，输出为这一范围内奇数的总数。

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 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

Sample Output

Author

HIT

Source

2012 Multi-University Training Contest 5

http://blog.youkuaiyun.com/sky1203850702/article/details/7844035 博客

#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;
        }
        long long ans = 1;
        for(int i = 0;i < res;i++)
            ans*=2;
        cout<<ans<<endl;
    }
    return 0;
}