HDU 4349 Xiao Ming's Hope（数学题）

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最新推荐文章于 2025-08-17 21:38:20 发布

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文章标签： java

本文介绍了一种快速计算特定范围内奇数组合数数量的方法。通过分析二进制表示中的1的数量来确定奇数组合数的总数，提供了一个简洁高效的C语言实现。

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Xiao Ming's Hope

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 433 Accepted Submission(s): 290

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
  

Source

2012 Multi-University Training Contest 5

Recommend

zhuyuanchen520

求 C _(n,0) ,C _(n,1) ,C _(n,2) ...C _(n,n) .当中有多少个奇数。

就是统计N表示成二进制有多少个1，然后就是就是2^cnt。

#include<stdio.h>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int cnt=0;
        while(n)
        {
            if(n&1)cnt++;
            n>>=1;
        }
        printf("%d\n",1<<cnt);
    }
    return 0;
}