HDU 4349——Xiao Ming's Hope——————【Lucas定理】

Xiao Ming's Hope

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1786    Accepted Submission(s): 1182

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

 

 

题目大意：给你一个数n，让你求C(n,0)、C(n,1)...C(n,n)这n+1个数中为奇数的个数。

解题思路：用Lucas定理。Lucas定理是用来求 c(n,m) mod p，p是素数的值。我们将n化成二进制串。C(A,B)≡C(a[n],b[n])*C(a[n-1],b[n-1])*C(a[n-2],b[n-2])*...C(a[0]*b[0])%p。这里p是2。如果A为10010。B从0 -> 10010枚举。C(0,1)为0。如果n的二进制串中该位置为0，那么要让C(A,B)%2==1那么，只能让m的二进制对应位置为0，对于n的二进制中为1的位置，m的二进制对应位置为0或1的结果都是1。所以结果就是n的二进制中1的位置取2或1的所有可能。即2^k，k为n的二进制中1的个数。

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        int sum=0;
        while(n){
            if(n&1)
                sum++;
            n>>=1;
        }
        printf("%d\n",(int)pow(2,sum));
    }
    return 0;
}