Xiao Ming's Hope HDU - 4349(Lucas定理)

   

Xiao Ming's Hope

HDU - 4349
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

这道题需要使用卢卡斯定理，什么事卢卡斯定理呢 下面只是简单说一下什么事卢卡斯定理和代码实现，不做证明

首先，Lucas（卢卡斯）定理是什么？有什么用？

Lucas定理是用来求 C(n,m) mod p，p为素数的值。（注意：p一定是素数）

Lucas定理用来解决大组合数求模是很有用的。

For non-negative integers m and n and a prime p, the following congruence relation holds:

where

and

are the base  p  expansions of  m  and  n  respectively.

即变成位运算更简单点说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，数组里面存的是二进制位 递归方程：(C(n%p, m%p)*Lucas(n/p, m/p))%p。（递归出口为m==0，return 1）

实现：

    #include<bits/stdc++.h>  
    using namespace std;  
    typedef long long ll;  
    const int N =1e5;  
    ll n, m, p, fac[N];  
    void init()  
    {  
        int i;  
        fac[0] =1;  
        for(i =1; i <= p; i++)  
            fac[i] = fac[i-1]*i % p;  
    }  
    ll q_pow(ll a, ll b)  
    {  
        ll  ans =1;  
        while(b)  
        {  
            if(b &1)  ans = ans * a % p;  
            b>>=1;  
            a = a*a % p;     
        }  
        return  ans;  
    }  
      
    ll C(ll n, ll m)  
    {  
        if(m > n)  return 0;  
        return  fac[n]*q_pow(fac[m]*fac[n-m], p-2) % p;  
    }  
      
    ll Lucas(ll n, ll m )  
    {  
        if(m ==0)  return 1;  
        else return  (C(n%p, m%p)*Lucas(n/p, m/p))%p;  
    }  
      
    int main()  
    {  
        int t;  
        scanf("%d", &t);  
        while(t--)  
        {  
            scanf("%I64d%I64d%I64d", &n, &m, &p);  
            init();  
            printf("%I64d\n", Lucas(n, m));  
        }  
        return 0;  
    }  

下面是这道题的解释和代码： 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。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main(){
    int n;
    while(~scanf("%d",&n)){
        int sum = 0;
        while(n){
            if(n&1){
                sum++;
            }
            n >>= 1;
        }
        printf("%d\n",(int)pow(2,sum));
    }
    return 0;
}