C - Xiao Ming's Hope----lucas定理

最新推荐文章于 2018-11-05 18:30:51 发布

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本文介绍了一道经典的数学问题，即计算特定范围内组合数中奇数值的数量。通过使用Lucas定理，文章提供了一个高效的解决方案，展示了如何将整数转换为二进制并利用位操作来快速计算结果。

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C - 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

题目链接：https://cn.vjudge.net/contest/161625#problem/C

比赛的时候是打表找的规律，并没有想起来lucas定理，学了lucas定理好久了，有点忘了，我看了大神的这个lucas的解题报告，好吧，确实忘了lucas了。

以下来自大神：

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。

代码：

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