1011_xiao_ming's _ hope

Question:

Hdu_4349_xiao_ming's _ hope

 

Official question: In this article, we will give a rough analysis of C (n, m) % 2, so we can write the Lucas theorem into binary format for observation, such as N = 1001101, M is an enumeration from 000000 to 1001101. In this theorem, C () = 0, therefore, if the M binary bit of the corresponding position of 0 of N = 1001101 is 1, then C (n, m) % 2 = 0. Therefore, the position of M corresponding to N is 0 and only 0 can be entered, and fill in 0 in the position of 1, fill in 1 All 1 (C () = 1), do not affect the result is an odd number, and ensure that N is not out of the range, therefore, all the situations are the enumeration of 1 in N corresponding to 0, 1 in m, and the result is obviously: 2 ^ (number of 1 in N)

 

My personal understanding: As the question solves, this question is a direct application of the Lucas theorem. Of course, you can also write a few more answers to see that the relationship between the answer and N is 2 ^ (the number of 1 in N ). Lucas theorem:

For non-negative integersMAndNAnd a primeP, The following congruence relation holds:

Where

And

Are the basePExpansionsMAndNRespectively.

The theorem is not proven yet.

 

Bidding Process:

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>using namespace std;int lowbit(int x) {    return x & (-x);}int get_result(int n) {    int res = 0;    while (n) {        n -= lowbit(n);        res++;    }    return res;}int main() { //   freopen("data.in", "r", stdin); //   freopen("data.out", "w", stdout);    int n;    while (scanf("%d", &n) != EOF) {        printf("%d\n", 1 << get_result(n));    }    return 0;}