HDU 4349 Xiao Ming's Hope 简单的数学—组合数性质

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?

题意很简单，就是要判断组合数的奇偶。 由组合数的性质，c(n,m) 如果 n&m不为0 则为奇数

故对于一个n只要得到其二进制表示下1的个数 t  答案就是 2^t。

为了加快数度可以用快速幂

写了这篇过几天才知道这个是Lucas定理，和别人一比自己的代码简直挫爆了

#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define N 30100;
int cnt;
int n;
int cal(int x)
{
int ret = 0;
while(x != 0)
{
if(x&1)
ret++;
x >>= 1;
}
return ret;
}
int fast(int c,int time)
{
int ret = 1;
while(time)
{
if(time&1)
ret*=c;
c*=c;
time >>=1;
}
return ret;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
int time = cal(n);
// printf("%d\n",time);
int ans = fast(2,time);
printf("%d\n",ans);
}
return 0;
}