Xiao Ming's Hope
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2609 Accepted Submission(s): 1790
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
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算法分析：
题意：
给n，求c（n,0），c（n,1）....c（n,n）中奇数的个数
分析：
先给一个Lucas推论：
将n，m(0~n) 化为二进制数
a[],b[]为n,m的二进制数
根据定理 C（n,m）=C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0])
官方题解：
Lucas定理推导，我们分析一下 C(n,m)%2,那么由lucas定理，我们可以写成二进制的形式观察，比如 n=1001101，m是从000000到1001101的枚举，我们知道在该定理中C(0,1)=0,因此如果n=1001101的0对应位置的m二进制位为1那么C(n,m) % 2==0,因此m对应n为0的位置只能填0，而1的位置填0，填1都是1（C(1,0)=C(1,1)=1)，不影响结果为奇数，并且保证不会出n的范围，因此所有的情况即是n中1位置对应m位置0，1的枚举，那么结果很明显就是：2^(n中1的个数)。
规律：
判断组合数奇偶性的一个规律：如果(n&m)==m，那么C(n,m)为奇数，否则为偶数
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int sum=0;
while(n)
{
if(n&1) sum++;
n=n>>1;
}
int ans=(int)pow(2.0,sum*1.0);
cout<<ans<<endl;
}
return 0;
}