HDU4349-Xiao Ming's Hope(Lucas定理)

小咪咪 2022-06-06 12:22 119阅读 0赞

# Xiao Ming's Hope #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 2435    Accepted Submission(s): 1674**

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<=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

Author

HIT

Source

[2012 Multi-University Training Contest 5][]

Recommend

zhuyuanchen520

lucas定理：

A、B是非负整数，p是质数。AB写成p进制：A=a\[n\]a\[n-1\]...a\[0\]，B=b\[n\]b\[n-1\]...b\[0\]。  
则组合数C(A,B)与C(a\[n\],b\[n\])\*C(a\[n-1\],b\[n-1\])\*...\*C(a\[0\],b\[0\])  modp同余

即：Lucas(n,m,p)=c(n%p,m%p)\*Lucas(n/p,m/p,p)

题意：求C(n, m)中奇数的各数，用lucas定理可以将n，m转化为二进制， C(n, m)%2==1 也就是C(a\[n\],b\[n\])\*C(a\[n-1\],b\[n-1\])\*...\*C(a\[0\],b\[0\]) %1==1。因为C(0，1)==0，C(0, 0)==1, 也就是说，n的二进制位为“0”，m的二进制相应的那一位也必须是“0”，否则C(a\[n\],b\[n\])\*C(a\[n-1\],b\[n-1\])\*...\*C(a\[0\],b\[0\]) %1==0。而对于n二进制为“1”的位置来说m二进制位任意。因为C(1, 0)=C(1,1)==1.所以 只要判断n的二进制有多少个“1”，就可以求出m有多少种情况。结论就是2^(n的中1的个数)

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

[2012 Multi-University Training Contest 5]: http://acm.hdu.edu.cn/search.php?field=problem&key=2012+Multi-University+Training+Contest+5&source=1&searchmode=source